From L1 Closeness to Uniform Tail Control
Uniform integrability controls the part of a function that lies above a large magnitude threshold, uniformly across a family. L1 convergence gives more than bounded L1 norms: all sufficiently late sequence members are close in integral to one fixed integrable function. That common limit allows their tails to be controlled together. The proof makes this mechanism quantitative, then handles the finitely many early terms separately.
Throughout, let \((X,\mathcal{M},\mu)\) be a finite measure space, meaning \(\mu(X)<\infty\). The functions in this tutorial are measurable, real-valued functions, considered up to equality \(\mu\)-a.e. An integrable function \(g\) satisfies \(\int_X|g|\,d\mu<\infty\), and its \(L^1\) norm is \(\|g\|_1=\int_X|g|\,d\mu\). No completeness assumption on the measure space is needed.
Suppose \(f_n\to f\) in \(L^1(\mu)\), which means \(\|f_n-f\|_1\to0\). In particular, \(f\) and every \(f_n\) are integrable. The central estimate below compares the tail of \(f_n\) with two quantities: its distance from \(f\), and the tail of \(f\). We use the elementary integral bound \(\mu(\{|g|>a\})\le a^{-1}\int_X|g|\,d\mu\) for \(a>0\); it follows by integrating the pointwise inequality \(a\chi_{\{|g|>a\}}\le |g|\).
Proof. Let \(A=\{|u|>K\}\), and split \(A\) into \(A\cap\{|v|>K/2\}\) and \(A\cap\{|v|\le K/2\}\). On the first set, \(|u|\le |u-v|+|v|\), so its contribution is at most \(\|u-v\|_1+\int_X|v|\chi_{\{|v|>K/2\}}\,d\mu\). On the second set, the reverse triangle inequality gives \(|u-v|\ge\bigl||u|-|v|\bigr|>K/2\). Also \(|u|\le |u-v|+|v|\) there. The integral of the first term is at most \(\|u-v\|_1\). For the second term, \(|v|\le K/2\), and the set is contained in \(\{|u-v|>K/2\}\); hence $$ \int_{A\cap\{|v|\le K/2\}}|v|\,d\mu \le \frac K2\,\mu(\{|u-v|>K/2\}) \le \|u-v\|_1. $$ Adding the bounds proves the inequality. \(\square\)
The L1 Convergence Theorem
Proof. Fix \(\eta>0\). Since \(\|f_n-f\|_1\to0\), choose \(N\) such that \(\|f_n-f\|_1<\eta/8\) for every \(n\ge N\). The tail of \(f\) tends to zero: as the integer \(r\to\infty\), the functions \(|f|\chi_{\{|f|>r\}}\) converge pointwise to zero wherever \(f\) is finite, and are bounded by the integrable function \(|f|\). The Dominated Convergence Theorem therefore gives \(\int_X|f|\chi_{\{|f|>r\}}\,d\mu\to0\). Its hypotheses hold because \(f\) is finite-valued and integrable. Choose an integer \(r\) so large that this integral is less than \(\eta/4\), and set \(K=2r\).
For every \(n\ge N\), apply the lemma with \(u=f_n\) and \(v=f\). Then $$ \int_X|f_n|\chi_{\{|f_n|>K\}}\,d\mu \le 2\|f_n-f\|_1+\int_X|f|\chi_{\{|f|>r\}}\,d\mu <\frac{\eta}{4}+\frac{\eta}{4}=\frac{\eta}{2}. $$ For each of the finitely many indices \(n<N\), the same Dominated Convergence Theorem argument, now dominated by the integrable function \(|f_n|\), shows that \(\int_X|f_n|\chi_{\{|f_n|>K'\}}\,d\mu\to0\) as \(K'\to\infty\). Increase \(K\), if necessary, so every one of these finitely many integrals is less than \(\eta/2\). Increasing the threshold cannot increase any tail integral, so the bound for \(n\ge N\) remains valid. Thus every \(n\) has tail integral less than \(\eta/2\), and their supremum is at most \(\eta/2<\eta\). Since \(\eta\) was arbitrary, the required uniform limit is zero. \(\square\)
The order of the choices matters: first select \(N\) so the L1 errors are small, then select a threshold controlling the limit’s tail and the finitely many early terms. The lemma applies all late terms at that same threshold. Choosing a different threshold for each \(n\) would not establish uniform integrability.
Worked Example: A Bounded Sequence Converging in L1
On \([0,1]\) with Lebesgue measure \(m\), let \(f_n(x)=1+x/n\) and \(f(x)=1\). Compute the L1 error and a tail integral at the fixed threshold \(K=3/2\).
For \(0\le x\le1\), \(|f_n(x)-f(x)|=x/n\), so $$ \|f_n-f\|_1=\int_0^1\frac{x}{n}\,dx=\frac{1}{2n}. $$ This tends to zero, proving L1 convergence. For \(n=1\), \(f_1(x)=1+x\), and \(f_1(x)>3/2\) exactly when \(x>1/2\). Therefore $$ \int_0^1|f_1|\chi_{\{|f_1|>3/2\}}\,dm =\int_{1/2}^{1}(1+x)\,dx =\left[x+\frac{x^2}{2}\right]_{1/2}^{1} =\frac{7}{8}. $$ For every \(n\ge2\), \(f_n(x)\le1+1/n\le3/2\), so the strict tail event is empty and its integral is zero. Thus the supremum at this threshold is \(7/8\). For \(K\ge2\), all tail events are empty because \(f_n(x)\le2\). This verifies uniform tail control directly. As a check on the L1 calculation, \(n=2\) gives error \(1/4\), equal to \(\int_0^1 x/2\,dx\).
Worked Example: Convergent Spikes with Explicit Tail Bounds
On \([0,1]\) with Lebesgue measure, set \(f_n=n\chi_{(0,n^{-3})}\) and \(f=0\). Check L1 convergence and compute the supremum of the tail integrals for any \(K>0\).
The interval \((0,n^{-3})\) has measure \(n^{-3}\). Hence $$ \|f_n-f\|_1=\int_0^1|f_n|\,dm=n\cdot n^{-3}=\frac{1}{n^2}\longrightarrow0. $$ For a fixed \(K>0\), \(f_n\) takes only the values \(0\) and \(n\). Its strict tail event \(\{|f_n|>K\}\) is \((0,n^{-3})\) when \(n>K\), and is empty when \(n\le K\). Consequently, $$ \int_0^1|f_n|\chi_{\{|f_n|>K\}}\,dm =\begin{cases} 1/n^2,&n>K,\\ 0,&n\le K. \end{cases} $$ The smallest positive integer greater than \(K\) is \(\lfloor K\rfloor+1\), so the supremum over \(n\ge1\) is \(1/(\lfloor K\rfloor+1)^2\). For example, at \(K=2.5\), the first contributing index is \(n=3\) and the supremum is \(1/9\). This tends to zero as \(K\to\infty\), as the theorem predicts. The computation also shows why shrinking support by itself is not the key: the tail mass is controlled here because the L1 error tends to zero.
What Finiteness Adds—and a Common Mistake
On a finite measure space, the tail definition also implies a uniform bound on the L1 norms. Indeed, uniform integrability gives a threshold \(K_0\) for which every tail integral is at most \(1\). Splitting each integral over \(\{|f_n|\le K_0\}\) and \(\{|f_n|>K_0\}\) yields $$ \|f_n\|_1 \le K_0\mu(X)+1 $$ for every \(n\). The first term is finite because \(\mu(X)<\infty\). In fact, the proof of the theorem’s tail estimate uses L1 convergence and integrability of the limit; finite total measure is what makes this tail formulation also give bounded L1 norms.
Proof of the bound. On \(\{|f_n|\le K_0\}\), the integrand \(|f_n|\) is at most \(K_0\), so its integral is at most \(K_0\mu(X)\). The integral on the complementary strict tail set is at most \(1\) by the choice of \(K_0\). Adding these two estimates proves the displayed bound. \(\square\)
A common mistake is to infer uniform integrability from bounded L1 norms alone. On \([0,1]\) with Lebesgue measure, consider \(g_n=n\chi_{(0,1/n)}\). Every norm is exactly $$ \|g_n\|_1=n\cdot\frac1n=1. $$ But for each fixed \(K>0\), the strict tail event is \((0,1/n)\) whenever \(n>K\), and is empty otherwise. Thus its tail integral is \(1\) for \(n>K\), and the supremum over \(n\) is \(1\), not a quantity tending to zero. The error in the claim “the supports shrink, so the tails vanish” is that the height grows at exactly the rate needed to keep the tail integral equal to \(1\). This sequence does not converge in L1 to zero: its L1 distance from zero is \(1\) for every \(n\). The convergence hypothesis in the theorem cannot be replaced by boundedness of the L1 norms.
The result is useful because L1 convergence supplies a common integrable reference function and small errors for every sufficiently late term. The lemma turns those facts into a tail estimate at a single threshold; the finite initial segment can then be controlled individually. This is the mechanism by which L1 convergence ensures the uniform tail condition used in convergence arguments.
Check Your Understanding
Use the strict tail event \(\{|f_n|>K\}\) in each calculation.
- In the first worked example, compute the tail integral for \(n=1\) at the threshold \(K=1\).
- For the spike sequence \(f_n=n\chi_{(0,n^{-3})}\), what is the tail integral at \(K=4\) when \(n=4\), and when \(n=5\)?
- Which two terms control the late-index tail in the lemma when it is applied to \(f_n\) and \(f\)?
- Why does the proof control the finitely many indices \(n<N\) separately?
- For \(g_n=n\chi_{(0,1/n)}\), compute the L1 norm and the supremum of the tail integrals at an arbitrary fixed \(K>0\). Why does this not contradict the theorem?