Why Truncation Separates the Two Difficulties
Convergence in measure says that the set where two functions differ by more than a fixed positive amount becomes small. It does not, by itself, prevent a function from taking very large values on that small set. Uniform integrability addresses precisely those large values. Truncation combines the two ideas: first remove the large-value tails, then estimate the remaining bounded functions on the exceptional set.
Throughout, \((X,\mathcal{M},\mu)\) is a measure space with \(\mu(X)<\infty\). The functions in the theorem and examples are real-valued and measurable; integrability means \(\int_X |f|\,d\mu<\infty\). Convergence \(f_n\to f\) in measure means that for every \(\varepsilon>0\), \(\mu(\{|f_n-f|>\varepsilon\})\to0\). All L1 convergence statements mean \(\int_X|f_n-f|\,d\mu\to0\).
Vitali’s Theorem by Truncation
Proof. We first obtain a subsequence converging to \(f\) almost everywhere, so that Fatou’s lemma can be used for the tails of \(f\). For each integer \(j\ge1\), convergence in measure lets us choose indices \(n_j\), strictly increasing, such that $$ \mu\bigl(\{|f_{n_j}-f|>2^{-j}\}\bigr)<2^{-j}. $$ Write \(A_j=\{|f_{n_j}-f|>2^{-j}\}\). For every \(N\), countable subadditivity gives \(\mu(\bigcup_{j\ge N}A_j)\le\sum_{j\ge N}2^{-j}=2^{1-N}\). The sets \(\bigcup_{j\ge N}A_j\) decrease as \(N\) increases. Since \(\mu(X)<\infty\), continuity of measure from above gives \(\mu(\bigcap_N\bigcup_{j\ge N}A_j)=0\). Outside this null set, a point belongs to only finitely many \(A_j\), so \(|f_{n_j}-f|\le2^{-j}\) eventually. Thus \(f_{n_j}\to f\) almost everywhere.
Fix \(K>0\), and define \(r_K(t)=(|t|-K)_+=\max\{|t|-K,0\}\) for real \(t\). This is a nonnegative continuous function, so \(r_K(f_{n_j})\to r_K(f)\) almost everywhere. Fatou’s lemma for nonnegative measurable functions gives $$ \int_X r_K(f)\,d\mu\le\liminf_{j\to\infty}\int_X r_K(f_{n_j})\,d\mu \le\sup_n\int_{\{|f_n|>K\}}|f_n|\,d\mu. $$ The last inequality holds because \(r_K(f_n)=0\) where \(|f_n|\le K\), and \(r_K(f_n)\le|f_n|\) where \(|f_n|>K\). Denote the supremum on the right by \(\eta(K)\); uniform integrability says \(\eta(K)\to0\). Also, \(|f|\le K+r_K(f)\), so for any fixed \(K>0\), $$ \int_X|f|\,d\mu\le K\mu(X)+\eta(K)<\infty. $$ This proves that \(f\) is integrable.
For \(K>0\), let \(T_K(t)=\max\{-K,\min\{t,K\}\}\), the truncation of \(t\) to the interval \([-K,K]\). For every real \(t\), \(|t-T_K(t)|=r_K(t)\). Choose \(K\) large enough that \(\eta(K)<\varepsilon/4\), where \(\varepsilon>0\) is arbitrary. The tail estimate just proved gives \(\int_X|f-T_K(f)|\,d\mu\le\eta(K)\); the same pointwise comparison with the tail integral gives \(\int_X|f_n-T_K(f_n)|\,d\mu\le\eta(K)\) for every \(n\).
Fix \(\delta>0\) and set \(A_n=\{|f_n-f|>\delta\}\). The map \(T_K\) is 1-Lipschitz: for real \(a,b\), clipping both numbers to \([-K,K]\) cannot increase their distance, so \(|T_K(a)-T_K(b)|\le|a-b|\). Also, \(|T_K(a)-T_K(b)|\le2K\). On \(A_n^c\), the first bound is at most \(\delta\); on \(A_n\), use the second bound. Integration therefore yields $$ \int_X|T_K(f_n)-T_K(f)|\,d\mu \le\delta\mu(X)+2K\mu(A_n). $$ For this fixed \(K\), choose \(\delta>0\) so that \(\delta\mu(X)<\varepsilon/4\); for instance \(\delta=\varepsilon/(8(\mu(X)+1))\) works, including when \(\mu(X)=0\). Since \(f_n\to f\) in measure, \(\mu(A_n)\to0\), so for all sufficiently large \(n\), \(2K\mu(A_n)<\varepsilon/4\). The triangle inequality now gives $$ \int_X|f_n-f|\,d\mu \le\int_X|f_n-T_K(f_n)|\,d\mu +\int_X|T_K(f_n)-T_K(f)|\,d\mu +\int_X|T_K(f)-f|\,d\mu <\varepsilon. $$ The three terms are bounded respectively by \(\varepsilon/4\), \(\varepsilon/2\), and \(\varepsilon/4\), with the middle bound strict for sufficiently large \(n\). This proves L1 convergence. \(\square\)
Worked Examples
Worked Example: A Bounded Sequence with a Moving Exceptional Set
On \(([0,1],\mathcal{L},m)\), let \(f(x)=x\) and \(f_n(x)=x+\chi_{(0,1/n)}(x)\), for integers \(n\ge1\). The functions are measurable and real-valued. If \(0<\varepsilon<1\), the set \(\{|f_n-f|>\varepsilon\}\) is exactly \((0,1/n)\), of measure \(1/n\). If \(\varepsilon\ge1\), it is empty because the difference is either \(0\) or \(1\), and the inequality defining the set is strict. Thus \(f_n\to f\) in measure.
For every \(x\in[0,1]\), \(0\le f_n(x)\le2\). Hence for every \(K\ge2\), the strict tail set \(\{|f_n|>K\}\) is empty for every \(n\), and the sequence is uniformly integrable. The L1 error is computed exactly: $$ \int_0^1|f_n-f|\,dm=\int_0^1\chi_{(0,1/n)}\,dm=\frac1n. $$ For example, at \(n=4\) the exceptional set has measure \(1/4\), and the integral error is \(1/4\). Check: the difference equals \(1\) on a set of measure \(1/n\) and \(0\) elsewhere, so its integral is indeed \(1\cdot(1/n)+0\cdot(1-1/n)=1/n\).
Worked Example: An Unbounded Limit with Uniformly Controlled Tails
On \(([0,1],\mathcal{L},m)\), define \(f(0)=0\), \(f(x)=x^{-1/2}\) for \(0<x\le1\), and \(f_n(0)=0\), \(f_n(x)=x^{-1/2}\chi_{[1/n,1]}(x)\) for \(0<x\le1\). For each fixed \(x>0\), \(x\ge1/n\) for all sufficiently large \(n\), so \(f_n(x)\to f(x)\). At \(x=0\) the functions also converge to \(f(0)\); hence convergence holds everywhere, and therefore in measure.
For \(K\ge1\), \(0\le f_n\le f\), so $$ \int_{\{|f_n|>K\}}|f_n|\,dm \le\int_{\{f>K\}}f\,dm =\int_0^{K^{-2}}x^{-1/2}\,dx =2K^{-1}. $$ Here \(f(x)>K\) exactly when \(0<x<K^{-2}\); the endpoint does not affect the integral. The bound \(2/K\) tends to zero, uniformly in \(n\), so the sequence is uniformly integrable. The exact L1 error is $$ \int_0^1|f_n-f|\,dm=\int_0^{1/n}x^{-1/2}\,dx=2n^{-1/2}. $$ At \(n=4\), this is \(2/\sqrt4=1\). Check: \(\int_0^1x^{-1/2}\,dx=2\), and the omitted initial interval \((0,1/4)\) contributes \(2\sqrt{1/4}=1\), as computed.
Worked Example: A Finite Counting Measure Space
Let \(X=\{1,2,3\}\), let \(\mathcal{M}\) be all subsets of \(X\), and let \(\mu\) be counting measure, so \(\mu(E)\) is the number of elements of \(E\). Define \(f_n(1)=1+1/n\), \(f_n(2)=2\), \(f_n(3)=0\), and \(f(1)=1\), \(f(2)=2\), \(f(3)=0\). For any \(\varepsilon>0\), the set \(\{|f_n-f|>\varepsilon\}\) is \(\{1\}\) if \(1/n>\varepsilon\), and empty if \(1/n\le\varepsilon\). Its measure is therefore eventually zero, proving convergence in measure.
All values of every \(f_n\) are at most \(2\), so for \(K\ge2\) the strict tail sets are empty. Thus the sequence is uniformly integrable. Counting measure gives $$ \int_X|f_n-f|\,d\mu =|f_n(1)-f(1)|+|f_n(2)-f(2)|+|f_n(3)-f(3)| =\frac1n+0+0=\frac1n. $$ At \(n=4\), the three summands are \(1/4,0,0\), so the integral is \(1/4\). Check: only the first coordinate differs, by exactly \(1/4\) when \(n=4\), in agreement with the sum.
Why the Bounded-Part Estimate Needs Both Terms
A common mistake is to argue that a small exceptional set makes the integral of a bounded difference small, without accounting for the difference outside that set. For example, on \(([0,1],\mathcal{L},m)\), take \(g(x)=1/10\), \(h(x)=0\), and \(\delta=1/5\). The strict exceptional set \(\{|g-h|>\delta\}\) is empty, but \(\int_0^1|g-h|\,dm=1/10\), not zero. The incorrect estimate \(\int|g-h|\le2K\,m(\{|g-h|>\delta\})\) omits the complement of the exceptional set.
The correct estimate for functions truncated to \([-K,K]\) is $$ \int_X|T_K(f_n)-T_K(f)|\,d\mu \le \delta\mu(X)+2K\mu(\{|f_n-f|>\delta\}). $$ The first term accounts for the small, but not necessarily zero, difference on the complement; the second controls the exceptional set. Convergence in measure makes the second term small for fixed \(K\) and \(\delta\), while finiteness of \(\mu(X)\) makes the first term small when \(\delta\) is chosen small.
Truncation is useful because it divides the L1 problem into two tasks with different hypotheses. Uniform integrability controls the discarded tails, including the tail of the limit, and convergence in measure controls the bounded functions that remain. Neither estimate replaces the other: the proof needs both before it can conclude convergence in L1.
Check Your Understanding
For each question, identify which part uses uniform integrability and which part uses convergence in measure.
- In the theorem’s proof, why is a subsequence converging almost everywhere useful when estimating the tail of \(f\)?
- For \(T_K(t)=\max\{-K,\min\{t,K\}\}\), verify that \(|t-T_K(t)|=(|t|-K)_+\).
- On a finite counting measure space, why does convergence in measure of a sequence to \(f\) imply that, for each fixed \(\delta>0\), the exceptional set is eventually empty?
- In the bounded-part estimate, where is the assumption \(\mu(X)<\infty\) used?
- Why does the estimate for \(\int r_K(f)\,d\mu\) not require \(f_n\to f\) almost everywhere along the entire sequence?