From a Difficult Integral to a Bounded Local Problem
When \(f_n\to f\) almost everywhere, convergence of the integrals does not follow by itself. A useful proof strategy is to separate the parts of the functions that are difficult to control from the part where a standard convergence theorem applies. First discard the region outside a finite-measure set; then discard values above a fixed threshold. If both discarded contributions are small uniformly in \(n\), the remaining functions are bounded on a finite-measure set, where the Dominated Convergence Theorem applies.
Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space; it is not assumed finite, sigma-finite, or complete. All functions in this tutorial are measurable and real-valued. A function \(h\) is integrable if \(\int_X |h|\,d\mu<\infty\). Convergence almost everywhere means that there is a measurable set of measure zero outside which pointwise convergence holds.
The truncation removes large values without changing values within \([-K,K]\). The size of the part removed is controlled by the integral of \(|h|\) where \(|h|>K\).
Proof. If \(|h(x)|\le K\), then \(T_K(h(x))=h(x)\), so both sides of the pointwise inequality are zero. If \(h(x)>K\), then \(T_K(h(x))=K\), and \(|h(x)-T_K(h(x))|=h(x)-K\le h(x)=|h(x)|\). If \(h(x)<-K\), then \(T_K(h(x))=-K\), and \(|h(x)-T_K(h(x))|=-K-h(x)\le -h(x)=|h(x)|\). These cases prove the pointwise inequality. Integrating it and using monotonicity of the Lebesgue integral gives the stated integral bound. \(\square\)
A Tail-Control Criterion
The criterion below makes the truncation strategy precise. Its tail bounds must hold uniformly over the whole sequence, and the spatial set \(E\) and value threshold \(K\) must be the same for every \(n\) at a given tolerance. The set and threshold may depend on that tolerance.
Proof. Fix \(\eta>0\), and use the tail assumptions with \(\delta=\eta/8\). This gives a single finite-measure set \(E\) and a single threshold \(K\) for which all four displayed tail bounds hold for every \(n\). Since \(T_K\) is continuous, \(T_K(f_n)\to T_K(f)\) \(\mu\)-a.e. on \(E\). Moreover, \(|\chi_E T_K(f_n)|\le K\chi_E\), and \(\int_X K\chi_E\,d\mu=K\mu(E)<\infty\). The Dominated Convergence Theorem, applied on \((X,\mathcal{M},\mu)\), therefore gives \[ \int_E T_K(f_n)\,d\mu\longrightarrow\int_E T_K(f)\,d\mu. \] In particular, for all sufficiently large \(n\), the absolute difference between these two integrals is less than \(\eta/2\).
For each \(n\), split the difference of the full integrals into spatial tails, value-truncation errors on \(E\), and the difference of the truncated integrals. The triangle inequality and the lemma give \[ \begin{aligned} \left|\int_X f_n\,d\mu-\int_X f\,d\mu\right| &\le \int_{E^c}|f_n|\,d\mu+\int_{E^c}|f|\,d\mu\\ &\quad+\int_E|f_n-T_K(f_n)|\,d\mu+\int_E|f-T_K(f)|\,d\mu\\ &\quad+\left|\int_E T_K(f_n)\,d\mu-\int_E T_K(f)\,d\mu\right|. \end{aligned} \] Each of the first four terms is at most \(\delta=\eta/8\): for the value-truncation terms this follows from the lemma and the assumed bounds on \(E\cap\{|f_n|>K\}\) and \(E\cap\{|f|>K\}\). Their sum is at most \(\eta/2\). For all sufficiently large \(n\), the last term is less than \(\eta/2\). Thus the full difference is less than \(\eta\), proving convergence of the integrals. \(\square\)
The theorem is a sufficient criterion, not a claim that these tail conditions are necessary in every setting. Its proof isolates the roles of its hypotheses: the spatial tails control what lies outside \(E\), the value tails control the truncation errors, and finite measure of \(E\) makes the constant bound \(K\) integrable. The Dominated Convergence Theorem is used only after these conditions have been checked.
Worked Applications
Worked Example: Cutting Off the Far End of the Half-Line
On \([0,\infty)\) with Lebesgue measure, let \(f_n(x)=e^{-x}\chi_{[0,n]}(x)\). For every fixed \(x\ge0\), \(f_n(x)\to f(x)=e^{-x}\). Both \(f_n\) and \(f\) are integrable, and \[ \int_0^\infty f_n(x)\,dx=\int_0^n e^{-x}\,dx=1-e^{-n},\qquad \int_0^\infty f(x)\,dx=1. \]
To use the criterion, choose \(E=[0,R]\), where \(R>0\). This set has finite measure \(R\). For every \(n\), \[ \int_{E^c}|f_n|\,dx\le\int_R^\infty e^{-x}\,dx=e^{-R}, \qquad \int_{E^c}|f|\,dx=e^{-R}. \] Also, \(0\le f_n,f\le1\), so with \(K=1\) both value-tail integrals over \(E\) are zero: the strict condition \(|f_n|>1\) (or \(|f|>1\)) never holds. Taking \(R\) large makes the spatial tails arbitrarily small, uniformly in \(n\). On \(E\), the truncated functions are bounded by \(1\), and the bound is integrable because \(m(E)=R\). The Dominated Convergence Theorem on \(E\) and the tail criterion give \(\int f_n\,dx\to\int f\,dx\). The exact integrals above check this conclusion directly, since \(1-e^{-n}\to1\).
Worked Example: Truncating Large Values Near the Origin
On \((0,1)\) with Lebesgue measure, define \[ f_n(x)=x^{-1/2}\chi_{[1/n,1]}(x). \] For every fixed \(x\in(0,1)\), once \(n\ge 1/x\) we have \(x\in[1/n,1]\), so \(f_n(x)\to f(x)=x^{-1/2}\). These functions are integrable, with \[ \int_0^1 f_n(x)\,dx=\int_{1/n}^1 x^{-1/2}\,dx =2\left(1-\frac{1}{\sqrt n}\right), \qquad \int_0^1 f(x)\,dx=2. \]
Take \(E=(0,1)\), which has finite measure. There is no spatial tail. For \(K\ge1\), the strict level set \(\{f_n>K\}\) is contained in \((0,K^{-2})\), since \(f_n(x)>K\) implies \(x^{-1/2}>K\), hence \(x<K^{-2}\). The same containment holds for \(\{f>K\}\). Therefore, for every \(n\), \[ \int_{\{f_n>K\}} f_n\,dx \le\int_0^{K^{-2}}x^{-1/2}\,dx=\frac{2}{K}, \qquad \int_{\{f>K\}} f\,dx =\frac{2}{K}. \] Choosing \(K\) sufficiently large makes both value tails as small as desired. On \(E\), the functions \(T_K(f_n)\) converge almost everywhere to \(T_K(f)\) and are bounded in absolute value by \(K\), which is integrable on \(E\). Dominated convergence and the tail criterion apply. For instance, at \(n=4\) the displayed integral is \(2(1-1/2)=1\); as \(n\to\infty\), it tends to \(2\), in agreement with the integral of \(f\).
Worked Example: Moving Spikes with Vanishing Total Mass
On \(\mathbb{R}\) with Lebesgue measure, let \(f_n(x)=n\chi_{(n,n+1/n^2)}(x)\). For any fixed \(x\), \(f_n(x)=0\) for all integers \(n>x\), so \(f_n(x)\to f(x)=0\). Each function is integrable, and its integral is its height times the length of its interval: \[ \int_{\mathbb{R}}f_n\,dx=n\left(\frac{1}{n^2}\right)=\frac1n\longrightarrow0=\int_{\mathbb{R}}f\,dx. \]
Here both spatial and value truncation are useful. Let \(R\ge2\) be an integer and \(E=[-R,R]\). If \(n<R\), then \(n+1/n^2\le n+1\le R\), so the support lies in \(E\). If \(n=R\), the support lies to the right of \(R\) and its integral is \(1/R\). If \(n>R\), its integral is \(1/n\le1/(R+1)\). Consequently \(\int_{E^c}|f_n|\,dx\le1/R\) for every \(n\), while the tail of \(f=0\) is zero.
For any \(K>0\), the strict value-tail set is empty when \(n\le K\). When \(n>K\), its integral on \(E\) is at most the integral over the whole support, namely \(1/n\le1/K\). Thus \(\int_{E\cap\{|f_n|>K\}}|f_n|\,dx\le1/K\) for every \(n\); the value tail of \(f\) is zero. Choose \(R\) and \(K\) large to meet the theorem's bounds. On \(E\), for every \(n>R\), \(f_n\) is identically zero, so the truncated functions converge there to zero and are bounded by \(K\). This verifies the local dominated-convergence step as well as both uniform tail controls.
This sequence has no nonnegative integrable function \(g\) that dominates every \(f_n\) almost everywhere. Indeed, if \(g\ge f_n\) almost everywhere for every \(n\), countability gives a common conull set on which all these inequalities hold. The intervals \((n,n+1/n^2)\) are disjoint, and on the \(n\)-th interval \(g\ge n\) almost everywhere. Hence \[ \int_{\mathbb{R}}g\,dx\ge \sum_{n=1}^\infty n\,m((n,n+1/n^2)) =\sum_{n=1}^\infty \frac1n=\infty, \] contradicting integrability. The tail criterion can therefore prove convergence even when this type of common integrable dominator is unavailable.
A Pitfall: Pointwise Convergence Does Not Control the Tails
On \((0,1)\) with Lebesgue measure, consider \(u_n=n\chi_{(0,1/n)}\). For each fixed \(x>0\), \(u_n(x)=0\) once \(n\ge1/x\), so \(u_n\to0\) pointwise. Nevertheless, \[ \int_0^1u_n\,dx=n\cdot\frac1n=1 \] for every \(n\), whereas the integral of the limit is zero. The missing control is visible in the value tails: for any fixed \(K>0\), whenever \(n>K\), the set \(\{u_n>K\}\) is \((0,1/n)\), and its contribution is \(\int_{\{u_n>K\}}u_n\,dx=1\). It cannot be made uniformly small by choosing \(K\) large.
There is also no nonnegative integrable \(g\) dominating every \(u_n\) almost everywhere. If such a \(g\) existed, countability would give a common conull set on which \(g\ge u_n\) for every \(n\). For each integer \(k\ge1\), on the shell \((1/(k+1),1/k)\) we have \(u_k=k\), so \(g\ge k\) almost everywhere on that shell. Its length is \(1/k-1/(k+1)=1/(k(k+1))\), and therefore the integral of \(g\) over it is at least \[ k\left(\frac1k-\frac1{k+1}\right)=\frac1{k+1}. \] The shells are disjoint, and \(\sum_{k=1}^\infty1/(k+1)=\infty\). Thus \(\int_0^1g\,dx=\infty\), a contradiction. The shell argument is essential: for each fixed \(x>0\), \(u_n(x)\) is eventually zero, not arbitrarily large. The contradiction comes from integrating the lower bounds over disjoint shells, not from a pointwise claim at a fixed \(x\).
In practice, check the proof in this order: identify a finite-measure set, bound the spatial tails uniformly, bound the value tails uniformly at one common threshold, and only then invoke dominated convergence on the truncated functions. If either family of discarded tails remains large, pointwise convergence on the retained region cannot repair the missing estimate.
Check Your Understanding
For each question, identify which truncation controls the discarded contribution and state why dominated convergence is valid wherever it is used.
- On \([0,\infty)\), let \(f_n=e^{-x}\chi_{[0,n]}\). Find the spatial tail bound outside \([0,R]\), uniformly in \(n\).
- For \(f_n(x)=x^{-1/2}\chi_{[1/n,1]}(x)\) on \((0,1)\), determine the value-tail integral for the strict set \(\{f_n>K\}\) when \(K\ge1\).
- For \(f_n=n\chi_{(n,n+1/n^2)}\) on \(\mathbb{R}\), verify the spatial tail bound outside \([-R,R]\) for integer \(R\ge2\).
- For \(u_n=n\chi_{(0,1/n)}\), compute the value-tail contribution when \(n>K\). Which uniform tail condition fails?
- In the theorem, why must the same finite-measure set and value threshold work for every \(n\) at a fixed tolerance?