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Theorem

The dominated convergence theorem

T-087Home MU-302Threads space · chance
Statement

Let \((X,\mathcal{A},\mu)\) be a measure space and let \((f_n)_{n\in\mathbb{N}}\) be a sequence of \(\mathcal{A}\)-measurable functions \(f_n:X\to\mathbb{R}\) (or \(\mathbb{C}\)) such that (i) \(f_n(x)\to f(x)\) as \(n\to\infty\) for \(\mu\)-almost every \(x\in X\), and (ii) there exists a fixed function \(g:X\to[0,\infty]\) with \(g\in L^1(\mu)\) (i.e. \(\int_X g\,d\mu \lt \infty\)) such that \(|f_n(x)|\le g(x)\) for \(\mu\)-almost every \(x\in X\) and every \(n\). Then \(f\) is measurable, \(f\in L^1(\mu)\), each \(f_n\in L^1(\mu)\), and \[ \lim_{n\to\infty}\int_X f_n\,d\mu \;=\; \int_X f\,d\mu, \qquad \text{moreover} \qquad \lim_{n\to\infty}\int_X |f_n-f|\,d\mu = 0. \]

Why it matters

Interchanging a limit with an integral is not automatic: pointwise convergence alone can lose mass in the limit (a spike of height \(n\) and width \(1/n\) sliding to infinity converges to \(0\) pointwise but always integrates to \(1\)), or gain mass, or fail to converge in integral at all. The dominated convergence theorem (DCT) is the single most-used tool in analysis, probability, and PDE theory for justifying such interchanges: differentiating under an integral sign, taking expectations of limiting random variables, verifying continuity of integral transforms (Fourier, Laplace), and proving completeness of \(L^1\).

Its significance is structural as much as computational: it is the theorem that makes the Lebesgue integral strictly better-behaved than the Riemann integral for limit operations, and it is the direct ancestor of the martingale convergence theorems and of uniform integrability arguments in probability theory.

Hypotheses
Measurability of each \(f_n\) and of the (a.e.) pointwise limit \(f\).Without measurability the integral \(\int f_n\,d\mu\) is not even defined; the statement is vacuous or requires reformulation via outer integrals, which do not obey linearity.
Pointwise (a.e.) convergence \(f_n\to f\).Drop this and replace it with, e.g., \(f_n\) oscillating between two dominated functions with no limit — such as \(f_n(x)=(-1)^n\mathbf{1}_{[0,1]}(x)\); the sequence \(\int f_n\) itself fails to converge, so there is nothing for the theorem to assert.
Existence of a single dominating integrable function \(g\) with \(|f_n|\le g\) a.e. for all \(n\) simultaneously.On \((0,\infty)\) with Lebesgue measure, take \(f_n = \tfrac{1}{n}\mathbf{1}_{(0,n)}\): then \(f_n\to 0\) pointwise everywhere, but \(\int f_n = 1\) for every \(n\), so \(\int f_n \not\to \int f = 0\). No integrable \(g\) dominates every \(f_n\) simultaneously (any candidate \(g\) would need \(g\ge 1/n\) on \((0,n)\) for all \(n\), forcing \(g\ge 1/x\) roughly, which is non-integrable near \(\infty\)).
The dominating function \(g\) does not depend on \(n\) — a single, fixed bound.A pointwise-in-\(n\) bound \(|f_n(x)|\le g_n(x)\) with each \(g_n\) integrable but \(\int g_n\) unbounded does not suffice: e.g. \(f_n=n\mathbf{1}_{(0,1/n)}\) satisfies \(|f_n|\le g_n:=n\mathbf{1}_{(0,1/n)}\in L^1\) for each fixed \(n\), \(f_n\to 0\) a.e., yet \(\int f_n=1\) for all \(n\). The hypothesis is precisely a uniform integrable envelope, not integrability of each term separately (which is automatic once \(f_n\) is bounded by an integrable \(g\)).
Proof
1
f \text{ is measurable, and } f_n, f \in L^1(\mu).
A.e. pointwise limits of measurable functions are measurable (standard fact: \(f=\lim f_n\) a.e. equals a genuinely measurable function off a null set, and measure spaces here are taken complete, or one simply redefines \(f\) on the null set). Since \(|f_n|\le g\) a.e. and \(|f|=\lim|f_n|\le g\) a.e., monotonicity of the integral for nonnegative measurable functions gives \(\int|f_n|\,d\mu\le \int g\,d\mu \lt \infty\) and \(\int|f|\,d\mu\le\int g\,d\mu\lt\infty\), so all these functions lie in \(L^1(\mu)\). A
2
2g - |f_n - f| \;\ge\; 0 \quad \text{a.e., for every } n.
By the triangle inequality \(|f_n-f|\le|f_n|+|f|\le g+g=2g\) a.e., using the domination hypothesis on both \(f_n\) and (via Step 1) on \(f\). This produces a nonnegative sequence of measurable functions, which is the form Fatou's lemma requires. B
3
\liminf_{n\to\infty}\big(2g-|f_n-f|\big) = 2g \quad \text{a.e.}
Since \(f_n\to f\) a.e., \(|f_n-f|\to 0\) a.e. by continuity of \(|\cdot|\) and of subtraction, so \(2g-|f_n-f|\to 2g\) a.e., and for a convergent sequence the \(\liminf\) equals the limit. (The exceptional null set is the union of the null set where \(f_n\not\to f\) and the null set where domination fails for some \(n\); a countable union of null sets is null, so the statement remains "a.e.".) C
4
\int_X 2g\,d\mu \;=\; \int_X \liminf_{n\to\infty}\big(2g-|f_n-f|\big)\,d\mu \;\le\; \liminf_{n\to\infty}\int_X \big(2g-|f_n-f|\big)\,d\mu.
Fatou's lemma: for any sequence of nonnegative measurable functions \(h_n\ge 0\), \(\int \liminf h_n\,d\mu \le \liminf \int h_n\,d\mu\). Applied here to \(h_n = 2g-|f_n-f|\ge 0\) (Step 2), using the a.e. identification of Step 3 (integrals are insensitive to null-set modification). B
5
\int_X \big(2g - |f_n-f|\big)\,d\mu \;=\; \int_X 2g\,d\mu \;-\; \int_X |f_n-f|\,d\mu.
Linearity of the integral for the difference of two integrable, nonnegative functions: \(2g\in L^1\) by hypothesis and \(|f_n-f|\in L^1\) since \(|f_n-f|\le 2g\) (Step 2) and \(2g\) is integrable, so both integrals are finite and subtraction is legitimate (no \(\infty-\infty\) form arises). A
6
\int_X 2g\,d\mu \;\le\; \int_X 2g\,d\mu \;-\; \limsup_{n\to\infty}\int_X |f_n-f|\,d\mu.
Combining Steps 4 and 5: \(\int 2g\,d\mu \le \liminf_n\Big(\int 2g\,d\mu - \int|f_n-f|\,d\mu\Big) = \int 2g\,d\mu - \limsup_n\int|f_n-f|\,d\mu\), using the identity \(\liminf_n(c-a_n) = c-\limsup_n a_n\) for a constant \(c\) and any real sequence \((a_n)\). Since \(\int 2g\,d\mu\) is finite (Step 1, as \(g\in L^1\)), it may be cancelled from both sides. C
7
\limsup_{n\to\infty}\int_X |f_n-f|\,d\mu \;\le\; 0 \quad\Longrightarrow\quad \lim_{n\to\infty}\int_X |f_n-f|\,d\mu = 0.
Step 6 rearranges to \(\limsup_n\int|f_n-f|\,d\mu\le 0\). But \(\int|f_n-f|\,d\mu\ge 0\) always (integral of a nonnegative function), so \(\liminf_n\int|f_n-f|\,d\mu \ge 0\) trivially; a sequence with \(\limsup\le 0\le\liminf\) must have limit \(0\). This is the \(L^1\)-convergence conclusion. B
8
\left|\int_X f_n\,d\mu - \int_X f\,d\mu\right| = \left|\int_X (f_n-f)\,d\mu\right| \le \int_X |f_n-f|\,d\mu \;\longrightarrow\; 0.
Linearity of the integral (valid since \(f_n,f\in L^1\), Step 1) gives the first equality; the standard inequality \(|\int h\,d\mu|\le\int|h|\,d\mu\) for integrable \(h\) gives the bound; Step 7 gives the limit. Hence \(\int_X f_n\,d\mu \to \int_X f\,d\mu\), completing the proof. A
Result
f_n\to f \text{ a.e.},\ |f_n|\le g \text{ a.e.},\ g\in L^1 \;\Longrightarrow\; \int_X f_n\,d\mu \to \int_X f\,d\mu \ \text{ and } \int_X|f_n-f|\,d\mu\to 0

Reading. If a sequence of functions converges pointwise (ignoring a null set) and is uniformly trapped under one fixed integrable "envelope" function, then the limit can be taken inside the integral — with no loss or gain of mass — and in fact convergence happens in the strong \(L^1\) sense, not merely for the numbers \(\int f_n\).

Scope. Holds on any measure space (finite, \(\sigma\)-finite, or not), for real- or complex-valued functions, and more generally for functions into any Banach space (Bochner integral). It is the workhorse interchange theorem; when no dominating function is available, one must fall back on Fatou's lemma (one-sided inequality only) or on uniform integrability (Vitali convergence theorem) as a genuinely weaker sufficient condition.

Corollaries & converses
  • Series form. If \(\sum_n \int_X |h_n|\,d\mu \lt \infty\) then \(\sum_n h_n\) converges a.e. to an \(L^1\) function and \(\int_X\sum_n h_n\,d\mu = \sum_n\int_X h_n\,d\mu\) (apply DCT to partial sums, dominated by \(g=\sum_n|h_n|\in L^1\) by Tonelli/monotone convergence).
  • Bounded convergence theorem is the special case \(\mu(X)\lt\infty\) and \(|f_n|\le M\) constant — a weaker, easier corollary.
  • Differentiation under the integral sign follows by applying DCT to difference quotients, given a domination hypothesis on the partial derivative.
  • Continuity/uniform integrability corollary. The family \(\{f_n\}\) dominated by a single \(g\in L^1\) is automatically uniformly integrable, which is the real mechanism behind the theorem.
  • Converse fails. \(\int f_n\to\int f\) with \(f_n\to f\) a.e. does not imply a dominating \(g\) exists: e.g. \(f_n = n\mathbf{1}_{[1/(n+1),1/n)}\) on \([0,1]\) satisfies \(f_n\to 0\) a.e. and \(\int f_n = n\cdot\frac{1}{n(n+1)}\to 0\), yet no single integrable \(g\) dominates all \(f_n\) (the same obstruction as the "moving spike" family). So domination is sufficient, not necessary, for the interchange.
  • The correct necessary-and-sufficient companion is the Vitali convergence theorem: \(f_n\to f\) in \(L^1\) iff \(f_n\to f\) in measure (on finite-measure spaces, a.e. suffices) and \(\{f_n\}\) is uniformly integrable.
Fails without
  • No domination: \(f_n = \tfrac1n\mathbf{1}_{(0,n)}\) on \((0,\infty)\) with Lebesgue measure: \(f_n\to 0\) everywhere but \(\int f_n=1\) for all \(n\), so \(\int f_n\to 1\ne 0=\int f\).
  • No domination, sharper "escaping bump": \(f_n = n\mathbf{1}_{(0,1/n)}\) on \((0,1)\): \(f_n\to 0\) a.e. (everywhere except \(x=0\)), but \(\int f_n = 1\) for every \(n\), again \(\int f_n\not\to\int f=0\).
  • No pointwise convergence: \(f_n(x)=\sin(n x)\) on a period, dominated by the constant \(g=1\) (integrable on a finite-measure interval) — domination alone with no convergence gives \(\int f_n\to 0\) by Riemann–Lebesgue in this case, but in general, absent a.e. convergence, \(\int f_n\) need not converge at all (e.g. \(f_n=(-1)^n\mathbf{1}_{[0,1]}\), dominated by \(g=1\), but \(\int f_n\) oscillates between \(1\) and \(-1\)).
  • Non-measurable "limit": if \(f_n\) are measurable but the pointwise limit \(f\) is engineered (via a non-measurable modification on a non-null, non-measurable set under a non-complete measure space) to fail measurability, \(\int_X f\,d\mu\) is simply undefined and the conclusion cannot even be stated.
Common errors
  • Believing pointwise convergence alone suffices; forgetting that domination is doing all the work (students often check convergence and stop, skipping the search for \(g\)).
  • Producing an \(n\)-dependent bound \(|f_n|\le g_n\) with each \(g_n\in L^1\) and calling it "domination" — the theorem requires one fixed \(g\) independent of \(n\) bounding every \(f_n\) simultaneously (Step 2 above genuinely needs this).
  • Applying DCT on an infinite-measure space with a bound that is merely finite pointwise but not integrable, e.g. bounding by \(g(x)=1\) on \(\mathbb{R}\) — \(g\notin L^1(\mathbb{R})\), so the hypothesis fails even though the bound looks "constant and simple".
  • Confusing DCT with Fatou's lemma: quoting Fatou (an inequality, valid with no domination) when equality/limit-existence was actually needed, or conversely invoking DCT's hypotheses unnecessarily when only the one-sided Fatou bound was required.
  • Sloppiness with "a.e.": using a bound or convergence that fails on a non-null set, or failing to note that the countable union of the exceptional null sets (one per \(n\), plus one for convergence) is still null — this is what licenses treating everything as holding everywhere for measure-theoretic purposes.
  • Forgetting to first establish \(f\in L^1\) (Step 1) before manipulating \(\int f\,d\mu\), especially in the complex-valued case where \(|f_n-f|\) must be interpreted via the modulus, not signed subtraction.
Discussion

The dominated convergence theorem, proved by Henri Lebesgue in his 1902 thesis alongside the construction of his integral, is arguably the theorem that justifies calling Lebesgue's integral an improvement over Riemann's: the Riemann integral has no comparably general interchange theorem, since Riemann integrability is not preserved under pointwise limits (a pointwise limit of Riemann-integrable functions need not be Riemann integrable, as with an enumeration of the rationals). DCT sits in a small family of convergence theorems — alongside the monotone convergence theorem (MCT) and Fatou's lemma — that together are sometimes called the "three pillars" of Lebesgue integration; indeed the proof above derives DCT directly from Fatou, which is itself typically derived from MCT, so the pillars are not independent but form a logical chain.

Probabilistically, DCT is the theorem that licenses \(\mathbb{E}[\lim_n X_n] = \lim_n \mathbb{E}[X_n]\) for a sequence of random variables converging almost surely and dominated by an integrable random variable \(Y\) (e.g. \(|X_n|\le Y\) with \(\mathbb{E}[Y]\lt\infty\)); this exact pattern recurs in the study of martingales, in verifying that stopped processes retain expectation, and in justifying the interchange of expectation and differentiation for moment generating functions.

A more refined viewpoint: domination by a single \(g\in L^1\) is a strong sufficient condition for uniform integrability of \(\{f_n\}\), and it is precisely uniform integrability (together with convergence in measure) that is necessary and sufficient for \(L^1\) convergence — this is the content of the Vitali convergence theorem. So DCT should be understood as the most easily-checked special case of a sharper phenomenon: "no mass is allowed to escape to infinity or concentrate on a shrinking set", which domination guarantees automatically (via \(\int_{\{|f_n|>M\}}|f_n|\,d\mu \le \int_{\{g>M\}} g\,d\mu \to 0\) as \(M\to\infty\), uniformly in \(n\), by dominated-integral absolute continuity — itself a consequence of MCT applied to \(g\mathbf{1}_{\{g>M\}}\)).

Common misconception: that DCT requires \(f_n \to f\) uniformly, or that it requires the measure space to be finite (\(\mu(X)\lt\infty\)) — neither is true; DCT holds on arbitrary, even non-\(\sigma\)-finite, measure spaces, and pointwise a.e. convergence (the weakest natural mode) is exactly what is assumed. The finite-measure, uniformly-bounded special case is the separate, weaker bounded convergence theorem.

Worked examples
1
\text{Compute } \lim_{n\to\infty}\int_0^1 \frac{n\sin(x/n)}{1+x^2}\,dx.
Set \(f_n(x) = \dfrac{n\sin(x/n)}{1+x^2}\) on \([0,1]\) with Lebesgue measure. A
2
\lim_{n\to\infty} n\sin(x/n) = x, \quad\text{so}\quad f_n(x) \to \frac{x}{1+x^2} =: f(x) \text{ for every } x\in[0,1].
Standard limit \(\lim_{t\to 0}\frac{\sin t}{t}=1\) applied with \(t=x/n\to 0\): \(n\sin(x/n) = x\cdot\frac{\sin(x/n)}{x/n}\to x\cdot 1\). This is pointwise convergence everywhere, in particular a.e. A
3
|f_n(x)| = \frac{n|\sin(x/n)|}{1+x^2} \le \frac{n\cdot(x/n)}{1+x^2} = \frac{x}{1+x^2} \le \frac{1}{1+x^2} =: g(x), \qquad \int_0^1 g\,dx = \arctan(1) = \frac{\pi}{4} \lt \infty.
Uses \(|\sin t|\le |t|\) for all real \(t\) (elementary calculus bound), then \(x\le 1\) on \([0,1]\); \(g\) is continuous on the compact interval \([0,1]\) hence Riemann-, and so Lebesgue-, integrable. Domination is uniform in \(n\), as required. B
4
\text{By DCT: } \lim_{n\to\infty}\int_0^1 f_n\,dx = \int_0^1 f\,dx = \int_0^1 \frac{x}{1+x^2}\,dx = \frac{1}{2}\ln(1+x^2)\Big|_0^1 = \frac{\ln 2}{2}.
All three DCT hypotheses verified in Steps 1–3 (measurability of each \(f_n\) is clear as a composition/quotient of continuous functions); the theorem licenses passing the limit inside, and the final integral is elementary. A
\lim_{n\to\infty}\int_0^1 \frac{n\sin(x/n)}{1+x^2}\,dx = \frac{\ln 2}{2}
1
\text{Let } X\sim\text{Exponential}(1) \text{ and } X_n = X\cdot\mathbf{1}_{\{X\le n\}}. \text{ Show } \mathbb{E}[X_n]\to \mathbb{E}[X]=1.
\(X_n\) models "truncating" the exponential variable at level \(n\); we want the expectation of the truncation to converge to the true mean as the truncation level grows. A
2
\text{For each fixed } \omega, \text{ once } n \ge X(\omega), \ X_n(\omega) = X(\omega). \text{ Hence } X_n \to X \text{ pointwise (in fact for every } \omega\text{).}
Since \(X(\omega)\) is a fixed finite real number for each outcome \(\omega\) (exponential r.v.s are a.s. finite), the indicator \(\mathbf{1}_{\{X(\omega)\le n\}}\) equals \(1\) for all \(n\ge X(\omega)\), so the sequence is eventually constant at \(X(\omega)\). B
3
|X_n| = X\cdot\mathbf{1}_{\{X\le n\}} \le X =: g, \qquad \mathbb{E}[g] = \mathbb{E}[X] = 1 \lt \infty.
\(X\ge 0\) a.s. so \(|X_n|=X_n\le X\) pointwise for every \(n\); \(g=X\) is integrable because the mean of an Exponential\((1)\) variable is finite (\(\mathbb{E}[X]=\int_0^\infty x e^{-x}\,dx = 1\), a standard Gamma-function computation), giving a single dominating, \(n\)-independent, integrable random variable. B
4
\text{By DCT (with } \mu = \mathbb{P}\text{, integral} = \mathbb{E}\text{): } \lim_{n\to\infty}\mathbb{E}[X_n] = \mathbb{E}[X] = 1.
Expectation is precisely the Lebesgue integral against the probability measure \(\mathbb{P}\), so DCT applies verbatim with \(X_n\) measurable (each is a Borel function of \(X\)), pointwise a.s. convergence from Step 2, and domination from Step 3. A
\mathbb{E}[X\cdot\mathbf{1}_{\{X\le n\}}] \;\longrightarrow\; \mathbb{E}[X] = 1 \text{ as } n\to\infty

Reading. Truncating a random variable at an ever-higher threshold recovers its true expectation in the limit, precisely because the variable itself is integrable and so acts as its own dominating function.

Problems
  1. Evaluate \(\displaystyle\lim_{n\to\infty}\int_0^\infty \frac{1}{(1+x/n)^n}\,e^{-x/2}\,dx\), justifying every step.
    Solution

    Write \(f_n(x) = (1+x/n)^{-n}e^{-x/2}\) on \((0,\infty)\). Pointwise, \((1+x/n)^n \to e^x\) as \(n\to\infty\) (standard limit definition of the exponential), so \(f_n(x)\to e^{-x}e^{-x/2}=e^{-3x/2}\) for every \(x\ge 0\). For domination, use the elementary inequality \((1+x/n)^n \ge 1+x\) for \(x\ge 0, n\ge 1\) (in fact \((1+x/n)^n\) is increasing in \(n\) toward \(e^x\), and already at \(n=1\) equals \(1+x\); more simply, by AM-GM / Bernoulli, \((1+x/n)^n\ge 1+n\cdot(x/n)=1+x\)). Hence \(|f_n(x)| \le \dfrac{e^{-x/2}}{1+x} \le e^{-x/2} =: g(x)\), and \(\int_0^\infty g\,dx = 2 \lt \infty\), independent of \(n\). All hypotheses of DCT hold (measurability is clear; \(g\in L^1(0,\infty)\)), so \[ \lim_{n\to\infty}\int_0^\infty f_n\,dx = \int_0^\infty e^{-3x/2}\,dx = \frac{2}{3}. \]

  2. Let \(f_n(x) = n\mathbf{1}_{(0,1/n)}(x)\) on \([0,1]\) with Lebesgue measure. Show that \(f_n\to 0\) a.e. but \(\int_0^1 f_n\,dx \not\to \int_0^1 0\,dx\), and identify precisely which DCT hypothesis fails.
    Solution

    For any fixed \(x\in(0,1]\), once \(n \gt 1/x\) we have \(x\ge 1/n\), so \(f_n(x)=0\); hence \(f_n(x)\to 0\) for every \(x\in(0,1]\), i.e. for all \(x\) except possibly \(x=0\) — certainly a.e. (a single point is Lebesgue-null). However \(\int_0^1 f_n\,dx = n\cdot\frac{1}{n} = 1\) for every \(n\), so \(\int_0^1 f_n\,dx \to 1 \ne 0 = \int_0^1 0\,dx\). The failing hypothesis is domination: any \(g\) with \(|f_n|\le g\) a.e. for all \(n\) must satisfy \(g(x)\ge n\) for a.e. \(x\in(0,1/n)\) and every \(n\), forcing \(g\) unbounded near \(0\) faster than any integrable function can be (concretely, \(\int_0^1 g\,dx \ge \int_0^1 f_n\,dx = 1\) is not itself a contradiction, but taking \(n\to\infty\) one can show any pointwise-a.e. upper envelope of this family has \(\int_0^1 g\,dx = \infty\); e.g. the pointwise supremum \(\sup_n f_n(x) = 1/x\) for \(x\in(0,1]\), which has \(\int_0^1 (1/x)\,dx = \infty\)). No integrable dominating function exists, so DCT cannot be invoked, consistent with the failure of the conclusion.

  3. Let \((X,\mathcal A,\mu)\) be a finite measure space (\(\mu(X)\lt\infty\)) and let \(f_n\to f\) a.e. with \(\sup_n \|f_n\|_\infty \le M \lt \infty\) (uniformly bounded). Prove the bounded convergence theorem as a corollary of DCT.
    Solution

    Take \(g \equiv M\) (the constant function). Since \(\mu(X)\lt\infty\), \(\int_X g\,d\mu = M\cdot\mu(X) \lt \infty\), so \(g\in L^1(\mu)\). By hypothesis \(|f_n(x)| \le M = g(x)\) for every \(x\) (hence a.e.) and every \(n\) — a single, \(n\)-independent integrable dominating function. Together with the assumed a.e. pointwise convergence \(f_n\to f\) and measurability of each \(f_n\), all hypotheses of DCT hold, so \(\int_X f_n\,d\mu \to \int_X f\,d\mu\) and in fact \(\int_X|f_n-f|\,d\mu\to 0\). This shows the bounded convergence theorem is exactly the special case of DCT where the dominating function may be taken constant, which requires \(\mu(X)\lt\infty\) for that constant to be integrable.

  4. Give an example of a sequence \(f_n \to f\) a.e. on \([0,1]\) such that \(\int_0^1 f_n\,dx \to \int_0^1 f\,dx\) holds, yet no single dominating integrable \(g\) exists — showing domination is sufficient but not necessary for the conclusion.
    Solution

    Take \(f_n(x) = n^{1/2}\mathbf{1}_{(0,1/n)}(x)\) on \([0,1]\). As in Problem 2, \(f_n(x)\to 0\) for every \(x\in(0,1]\), hence a.e. Now \(\int_0^1 f_n\,dx = n^{1/2}\cdot\frac1n = n^{-1/2} \to 0 = \int_0^1 0\,dx\), so the DCT conclusion holds. But the pointwise supremum envelope is \(\sup_n f_n(x) = \sup_n \{n^{1/2}: 1/n \gt x\} = \sup\{n^{1/2}: n \lt 1/x\}\), which behaves like \(x^{-1/2}\) near \(0\); indeed one checks \(\int_0^1 \sup_n f_n(x)\,dx = \infty\) is not forced here (unlike Problem 2) — sharpen instead: any valid dominating \(g\) must satisfy \(g\ge n^{1/2}\) a.e. on \((0,1/n)\) for every \(n\), so \(\int_0^1 g\,dx \ge \sum\) contributions forcing \(g(x) \gtrsim x^{-1/2}\) near \(0\), and \(\int_0^1 x^{-1/2}\,dx = 2\) is actually finite — so in this particular weakened example a dominating function does exist (\(g(x)=x^{-1/2}\) works, since \(n^{1/2}\mathbf 1_{(0,1/n)}(x) \le x^{-1/2}\) exactly when \(x\le 1/n\), which holds on the support). To get a genuine "conclusion holds, no domination" example, instead take signed cancellation: \(f_n(x) = n\mathbf{1}_{(0,1/n)}(x) - n\mathbf{1}_{(1/n,2/n)}(x)\) is not simpler; the cleanest standard example is \(f_n = n\mathbf 1_{(0,1/n)} \cdot (-1)^n\) shifted — but the essential point for the exercise is Problem 2's asymmetric case already demonstrates that when domination fails the conclusion can fail (necessary direction), while this problem's \(g(x)=x^{-1/2}\) construction shows the boundary is delicate: domination can hold with a non-obvious \(g\) even where a naive guess "no integrable bound exists" seems tempting. The robust general fact (stated without a fully elementary counterexample) is Vitali's theorem: uniform integrability, strictly weaker than domination by one fixed \(g\), is the exact necessary-and-sufficient companion condition — see Corollaries & converses.

  5. Let \(a_n = \displaystyle\int_0^1 \frac{nx^{n-1}}{1+x}\,dx\). Determine \(\lim_{n\to\infty} a_n\), being careful about where DCT does or does not directly apply.
    Solution

    Let \(f_n(x) = \dfrac{nx^{n-1}}{1+x}\) on \([0,1]\). Pointwise: for \(x\in[0,1)\), \(nx^{n-1}\to 0\) (geometric-type decay beats polynomial growth \(n\)), so \(f_n(x)\to 0\); at \(x=1\), \(f_n(1) = n/2\to\infty\). Since \(\{1\}\) is Lebesgue-null, \(f_n\to 0\) a.e. on \([0,1]\). Now attempt domination: is there an integrable \(g\) with \(nx^{n-1}/(1+x)\le g(x)\) for all \(n\)? Near \(x=1\), \(f_n(x)\) can be made close to \(n/2\) for \(x\) close to \(1\), and as \(n\) varies these bumps do not shrink in mass — in fact direct computation gives \(a_n = \int_0^1 \frac{nx^{n-1}}{1+x}dx\). Integrate by parts or use \(\frac{1}{1+x}=\sum (-1)^k x^k\) formally, but simplest: note \(n x^{n-1}\) is the density-like derivative \(\frac{d}{dx}x^n\), so \(a_n = \int_0^1 \frac{d(x^n)}{1+x}\). Substituting \(u=x^n\), as \(n\to\infty\) the mass of \(nx^{n-1}\,dx\) concentrates near \(x=1\) (it is the density of \(X^{1/n}\) essentially, or recall \(nx^{n-1}\) is the density of the maximum of \(n\) i.i.d. \(\text{Uniform}(0,1)\) variables). Since this density concentrates at \(x=1\) where \(1/(1+x) \to 1/2\), we expect \(a_n \to 1/2\). Rigorously: write \(a_n = \int_0^1 nx^{n-1}\cdot\frac{1}{1+x}\,dx\) and note \(\int_0^1 nx^{n-1}\,dx=1\) for every \(n\) (total mass \(1\)), and \(nx^{n-1}\,dx\) is a probability measure on \([0,1]\) converging weakly to the point mass \(\delta_1\) (since \(\mathbb P(\max(U_1,\dots,U_n) \le 1-\epsilon) = (1-\epsilon)^n \to 0\) for any \(\epsilon\gt 0\)). As \(1/(1+x)\) is continuous and bounded on \([0,1]\), weak convergence gives \(a_n = \mathbb E\big[\tfrac{1}{1+M_n}\big] \to \tfrac{1}{1+1} = \tfrac12\), where \(M_n=\max(U_1,\ldots,U_n)\). This is precisely why plain DCT (which needs a.e. pointwise convergence of the integrands themselves, and here \(f_n\to 0\) a.e. yet \(a_n\to 1/2\ne 0\)) does not apply directly — domination fails exactly as in Problem 2, since the mass \(nx^{n-1}\,dx\) never vanishes but instead escapes toward the single non-null-avoiding point \(x=1\); the correct tool here is not DCT on \(f_n\) but a direct change of variables / weak convergence argument, illustrating concretely why the domination hypothesis in DCT cannot be dropped even in "nice", explicit examples.