Dominated Convergence on a Finite Interval
On a bounded interval, a uniform bound is often enough to supply the integrable majorant required by the Dominated Convergence Theorem. The essential work is still to verify the other hypothesis: convergence almost everywhere to a measurable limit. Once both conditions hold, the integrals converge, even when convergence is not uniform and even when the values at a few endpoints behave differently.
Fix a finite interval \([a,b]\) with \(a<b\). In this tutorial, the measure space is \(([a,b],\mathcal{L}|_{[a,b]},m)\), where \(\mathcal{L}|_{[a,b]}\) is the Lebesgue sigma-algebra restricted to the interval and \(m\) is Lebesgue measure. This is a complete finite measure space, with \(m([a,b])=b-a\). Unless stated otherwise, functions are real-valued and measurable; limits and inequalities asserted almost everywhere may fail on a Lebesgue null set.
For a bounded interval there is a particularly convenient way to check the majorant. If \(|f_n(x)|\le M\) almost everywhere for every \(n\), where \(M\) is a finite constant, then \(g=M\chi_{[a,b]}\) is integrable and \(\int g\,dm=M(b-a)\). A bound depending on \(n\), such as \(|f_n|\le M_n\) with \(M_n\) unbounded, does not give the common majorant required by the theorem.
Proof. Remove the union of the null exceptional sets for the bounds and convergence. This is still a null set, since it is a countable union. At every remaining point, taking limits in \(|f_n|\le M\) gives \(|f|\le M\). Thus \(M\chi_{[a,b]}\) is an integrable majorant for \(f\) and for every \(f_n\). The Dominated Convergence Theorem gives integrability of \(f\) and convergence of the integrals. Also, \(f_n-f\to0\) almost everywhere and \(|f_n-f|\le |f_n|+|f|\le2M\) almost everywhere. Since \(2M\chi_{[a,b]}\) is integrable, another application of the Dominated Convergence Theorem gives \(\int_a^b|f_n-f|\,dm\to\int_a^b0\,dm=0\). This also covers \(M=0\), when all these functions vanish almost everywhere. \(\square\)
The proposition states convergence in \(L^1\), meaning that the \(L^1\)-norm \(\|f_n-f\|_1=\int_a^b|f_n-f|\,dm\) tends to zero. It does not assert uniform convergence: the values of \(f_n\) may remain far from their limits on small sets whose measures shrink to zero.
Three Applications
Worked Example: A Polynomial Sequence
For \(n\ge1\), let \(p_n(x)=x^2(1+x/n)\) on \([0,2]\). Find the pointwise limit and the limit of the integrals.
At each \(x\in[0,2]\), \(x/n\to0\), so \(p_n(x)\to x^2\). For every \(n\ge1\), \(0\le x/n\le2\), and therefore \(0\le p_n(x)\le3x^2\le12\). The constant function \(12\) is integrable on \([0,2]\), with integral \(12\cdot2=24\). Thus the hypotheses of the Dominated Convergence Theorem hold on the finite Lebesgue measure space \(([0,2],\mathcal{L}|_{[0,2]},m)\), and the limit integral is \(\int_0^2x^2\,dm\).
Check: the pointwise limit has integral \(\int_0^2x^2\,dx=[x^3/3]_0^2=8/3\), matching the limit computed from the sequence. The direct computation also shows that the difference between the \(n\)-th integral and its limit is exactly \(4/n\).
Worked Example: An Exponential Sequence Concentrating Near an Endpoint
For \(n\ge1\), let \(f_n(x)=e^{-nx}\) on \([0,1]\). Determine the almost-everywhere limit and the limit of the integrals.
At \(x=0\), \(f_n(0)=1\) for every \(n\). At each \(x>0\), \(e^{-nx}\to0\). Hence the pointwise limit is \(f=\chi_{\{0\}}\), which equals zero almost everywhere on \([0,1]\). Since \(0\le f_n\le1\) everywhere and \(m([0,1])=1\), the constant function \(1\) is an integrable majorant. Dominated convergence therefore gives \(\int_0^1 f_n\,dm\to\int_0^1 f\,dm=0\).
Check: because \(0<1-e^{-n}<1\), the displayed integral lies between \(0\) and \(1/n\), and so tends to zero. The limit function has integral zero because the singleton \(\{0\}\) has Lebesgue measure zero. The sequence does not converge uniformly: for every \(n\), its value at \(0\) is \(1\), whereas the limit at \(0\) is also \(1\), but at points approaching \(0\) from the right the difference \(e^{-nx}\) can be arbitrarily close to \(1\). Thus the supremum of the difference on \([0,1]\) is \(1\), not a quantity tending to zero.
Worked Example: Indicators of Shrinking Intervals
For integers \(n\ge2\), define \(E_n=[0,\tfrac12+\tfrac1n]\subseteq[0,1]\) and \(u_n=\chi_{E_n}\). Find the pointwise limit and show directly that the integrals converge to its integral.
If \(0\le x\le\tfrac12\), then \(x\in E_n\) for every \(n\), so \(u_n(x)=1\). If \(x>\tfrac12\), choose \(n\) sufficiently large that \(1/n<x-\tfrac12\); then \(x\notin E_n\), and \(u_n(x)=0\). It follows that \(u_n(x)\to\chi_{[0,1/2]}(x)\) at every \(x\in[0,1]\). Also \(0\le u_n\le1\), and \(1\) is integrable on this interval. Dominated convergence applies.
Check: the set on which \(u_n\) differs from \(\chi_{[0,1/2]}\) is \((1/2,1/2+1/n]\), whose measure is \(1/n\). Therefore \(\int_0^1|u_n-\chi_{[0,1/2]}|\,dm=1/n\), independently confirming convergence in \(L^1\). Whether an endpoint is included does not change these measures, since a singleton has measure zero.
What the Finite Interval Does—and Does Not—Provide
The finite measure of \([a,b]\) makes every constant bound integrable, but it does not make every measurable function integrable. A measurable function can be unbounded and fail to have a finite integral even on a bounded interval. In applications, the bound must be a single integrable function that works for all \(n\), not merely a separate finite bound for each function.
A common error is to check only the pointwise limit and then interchange limit and integral without verifying domination. On \([0,1]\), consider \(v_n=n\chi_{[0,1/n]}\). For each \(x>0\), \(v_n(x)\to0\), so \(v_n\to0\) almost everywhere. Nevertheless, direct computation gives \(\int_0^1v_n\,dm=n(1/n)=1\) for every \(n\), whereas the integral of the almost-everywhere limit is zero.
There cannot be an integrable common majorant for this sequence. Suppose a measurable \(g\ge0\) satisfied \(v_n\le g\) almost everywhere for every \(n\). Exclude the union of the exceptional null sets, one for each \(n\); outside this union all the inequalities hold simultaneously. For each integer \(k\ge1\), on \(I_k=(2^{-(k+1)},2^{-k}]\) we have \(x\le2^{-k}\), so \(v_{2^k}(x)=2^k\). Thus \(g\ge2^k\) almost everywhere on \(I_k\), and
The intervals \(I_k\) are disjoint. For every positive integer \(N\), additivity and monotonicity of the nonnegative integral give \(\int_0^1g\,dm\ge\sum_{k=1}^N\int_{I_k}g\,dm\ge N/2\). Since this holds for every \(N\), \(g\) cannot be integrable. The example fails the common-majorant hypothesis, so it does not contradict dominated convergence.
For a bounded-interval problem, a reliable check is to write down the interval and measure, identify the pointwise limit (including any exceptional endpoints), and then exhibit one integrable bound independent of \(n\). Only after those checks should the theorem be used to pass the limit through the integral.
Check Your Understanding
For each question, specify the relevant interval, pointwise or almost-everywhere limit, and the bound needed to justify any use of dominated convergence.
- On \([0,1]\), let \(q_n(x)=x^n\). Find its pointwise limit, verify an integrable common bound, and compute \(\lim_{n\to\infty}\int_0^1q_n\,dm\).
- On \([0,2]\), let \(r_n(x)=e^{-nx}\). Identify the limit at \(x=0\) and for \(x>0\), then compute the limit of the integrals.
- Let \(A_n=[0,\tfrac13+\tfrac1n]\subseteq[0,1]\) for \(n\ge2\). Find the pointwise limit of \(\chi_{A_n}\) and the measure of the set where it differs from that limit.
- Why does \(n\chi_{[0,1/n]}\) not satisfy the common integrable-majorant hypothesis, despite its almost-everywhere limit being zero?
- State the additional conclusion in \(L^1\) that follows when a uniformly bounded sequence on a finite interval converges almost everywhere.