From Pointwise Control to Small Error Sets
Almost-everywhere convergence is a pointwise statement outside a null set: each such point eventually has small error. Convergence in measure is a statement about the size of the set where the error exceeds a fixed threshold. On a finite measure space, these statements are linked because the points that have a large error at some time after \(N\) form a decreasing sequence of tail sets. Their intersection is null, and finiteness lets their measures decrease to zero.
Throughout the main result, \((X,\mathcal{M},\mu)\) is a measure space with \(\mu(X)<\infty\). The functions \(f_n,f:X\to\mathbb{R}\) are measurable. Almost-everywhere convergence means that \(f_n(x)\to f(x)\) for every \(x\) outside some measurable set of measure zero. Convergence in measure means that for every \(\varepsilon>0\), the measure of the strict error set \(\{x:|f_n(x)-f(x)|>\varepsilon\}\) tends to zero. Neither completeness nor sigma-finiteness is needed.
For fixed \(\varepsilon\), almost-everywhere convergence says that each point outside the exceptional null set belongs to only finitely many of the error sets \(E_n(\varepsilon)\). It does not say that the error sets themselves are decreasing. To use the pointwise information without making that assumption, take their tail unions.
The Finite-Measure Implication
Proof. Put \(A=\bigcap_{N=1}^{\infty}A_N\) and \(D_k=A_k\setminus A_{k+1}\). The sets \(A\) and \(D_k\), \(k\ge1\), are pairwise disjoint, and their union is \(A_1\): a point in \(A_1\setminus A\) leaves the decreasing sequence at exactly one index. Countable additivity gives \(\mu(A_1)=\mu(A)+\sum_{k=1}^{\infty}\mu(D_k)\). This sum is finite because \(\mu(A_1)<\infty\). For each \(N\), the same decomposition gives \(\mu(A_N)=\mu(A)+\sum_{k=N}^{\infty}\mu(D_k)\). The tails of a convergent series of nonnegative numbers tend to zero, so \(\mu(A_N)\to\mu(A)\). \(\square\)
Proof. Fix an arbitrary \(\varepsilon>0\) and set \(E_n=\{x:|f_n(x)-f(x)|>\varepsilon\}\). These sets are measurable because \(f_n-f\) is measurable. Define the tail unions \(B_N=\bigcup_{n\ge N}E_n\). They are measurable and satisfy \(B_1\supseteq B_2\supseteq\cdots\). Their intersection is the set of points that belong to infinitely many \(E_n\).
Let \(Z\in\mathcal{M}\) be a null set outside which \(f_n(x)\to f(x)\). If \(x\notin Z\), convergence gives an index \(N_x\) such that \(|f_n(x)-f(x)|\le\varepsilon\) for every \(n\ge N_x\). Thus \(x\notin E_n\) for every \(n\ge N_x\), so \(x\notin\bigcap_N B_N\). Consequently, \(\bigcap_N B_N\subseteq Z\), and \(\mu(\bigcap_N B_N)=0\).
Since \(B_1\subseteq X\), we have \(\mu(B_1)\le\mu(X)<\infty\). The lemma therefore applies and yields \(\mu(B_N)\to0\). For every \(n\ge N\), \(E_n\subseteq B_N\), and hence \(0\le\mu(E_n)\le\mu(B_N)\). Given any \(\eta>0\), choose \(N\) with \(\mu(B_N)<\eta\); then \(\mu(E_n)<\eta\) for all \(n\ge N\). This proves \(\mu(E_n)\to0\). Since the argument applies to every \(\varepsilon>0\), it proves convergence in measure. \(\square\)
The finite-measure hypothesis enters precisely in the estimate \(\mu(B_1)\le\mu(X)<\infty\). It ensures continuity from above is available for the tail unions. Almost-everywhere convergence still implies that their intersection is null on an infinite measure space, but that fact alone does not force the measures of the tail unions to tend to zero.
Worked Error-Set Calculations
Worked Example: Powers on the Unit Interval
Let \(X=[0,1]\) with Lebesgue measure \(m\), and let \(f_n(x)=x^n\). Show that \(f_n\to0\) \(m\)-a.e. and compute the error-set measure for each \(\varepsilon>0\).
For every \(x\in[0,1)\), \(x^n\to0\). At \(x=1\), the sequence tends to \(1\), so the pointwise limit agrees with \(f=0\) except on the null set \(\{1\}\). For \(0<\varepsilon<1\), the strict error condition \(x^n>\varepsilon\) is equivalent to \(x>\varepsilon^{1/n}\). Therefore \(E_n(\varepsilon)=(\varepsilon^{1/n},1]\), and \(m(E_n(\varepsilon))=1-\varepsilon^{1/n}\).
For example, at \(\varepsilon=\tfrac14\), the measures for \(n=1,2,4\) are \(1-\tfrac14=\tfrac34\), \(1-\tfrac12=\tfrac12\), and \(1-\tfrac14=\tfrac34\)? The last expression must use \((1/4)^{1/4}=1/\sqrt{2}\), so the measure at \(n=4\) is \(1-\tfrac1{\sqrt2}\). Thus the exact values are \(\tfrac34,\tfrac12,1-\tfrac1{\sqrt2}\), and the general formula, rather than monotonicity in \(n\), gives the limit. For \(\varepsilon\ge1\), the strict condition \(x^n>\varepsilon\) is never satisfied, so the error-set measure is zero.
Check: for \(0<\varepsilon<1\), \(\varepsilon^{1/n}\to1\), so the computed measure \(1-\varepsilon^{1/n}\) tends to zero as required.
Worked Example: Roots on the Unit Interval
Again take \(([0,1],\mathcal{L},m)\), and let \(g_n(x)=x^{1/n}\). The limit is \(g=1\) for \(x>0\), while \(g_n(0)=0\); hence \(g_n\to g\) \(m\)-a.e. For \(0<\varepsilon<1\), calculate the strict error set relative to \(g\).
For \(x>0\), \(1-x^{1/n}>\varepsilon\) is equivalent to \(x<(1-\varepsilon)^n\). The point \(x=0\) also satisfies the error inequality because \(|g_n(0)-g(0)|=1>\varepsilon\). Thus \(E_n(\varepsilon)=[0,(1-\varepsilon)^n)\), and \(m(E_n(\varepsilon))=(1-\varepsilon)^n\). At \(\varepsilon=\tfrac12\), substituting gives measures \((1/2)^1=1/2\), \((1/2)^2=1/4\), and \((1/2)^3=1/8\), which tend to zero. For \(\varepsilon\ge1\), the error is at most \(1\), so the strict error set is empty. This verifies convergence in measure directly.
Check: the endpoint \((1-\varepsilon)^n\) is excluded by the strict inequality, but a single endpoint has Lebesgue measure zero; the interval length is exactly \((1-\varepsilon)^n\).
Worked Example: Disjoint Error Sets and Their Tails
On \([0,1]\) with Lebesgue measure, set \(I_n=(2^{-n},2^{-(n-1)}]\), let \(h_n=\chi_{I_n}\), and take \(h=0\). The intervals are pairwise disjoint: adjacent intervals have a shared boundary value only as the upper endpoint of one, while the lower endpoint of the other is excluded. Every point belongs to at most one \(I_n\), so \(h_n(x)\to0\) at every \(x\).
For \(0<\varepsilon<1\), the strict error set is exactly \(I_n\), and its measure is \(2^{-(n-1)}-2^{-n}=2^{-n}\). For example, the measures at \(n=1,2,3\) are \(1/2,1/4,1/8\). The tail union is \(B_N=\bigcup_{n\ge N}I_n=(0,2^{-(N-1)}]\), so \(m(B_N)=2^{-(N-1)}\to0\). For \(\varepsilon\ge1\), the error set is empty because \(h_n\le1\) and the inequality defining it is strict.
Check: \(m(I_n)=2^{-n}\) tends to zero, and the tail-union computation gives \(m(B_N)=2\cdot2^{-N}=2^{-(N-1)}\), also tending to zero.
Why Finiteness Cannot Be Dropped
Worked Example: A Countable Space of Infinite Measure
Let \(X=\mathbb{N}\times\mathbb{N}\) with the counting measure \(\mu\), so a set has measure equal to its number of points, possibly \(+\infty\). Define \(q_n(i,j)=1\) when \(i=n\) and \(q_n(i,j)=0\) otherwise, and take \(q=0\). For each fixed \((i,j)\), \(q_n(i,j)\) is nonzero only when \(n=i\), so \(q_n(i,j)\to0\). This is pointwise, and therefore almost-everywhere, convergence.
For any \(0<\varepsilon<1\), the strict error set is \(E_n(\varepsilon)=\{n\}\times\mathbb{N}\). It contains infinitely many points, so \(\mu(E_n(\varepsilon))=+\infty\) for every \(n\); in particular, these measures do not tend to zero. For \(\varepsilon\ge1\), the strict error set is empty. The space has \(\mu(X)=+\infty\), and the failed step in the finite-space proof would be using \(\mu(B_1)<\infty\): here even \(B_1=X\) has infinite measure.
Check: each fixed point \((i,j)\) eventually has error zero, while at every index \(n\) the error set contains \((n,1),(n,2),\ldots\). Pointwise eventual control at each point does not bound the measure of the full error set on an infinite measure space.
A Common Pitfall
Do not assume that the error sets \(E_n(\varepsilon)\) decrease. In the disjoint-interval example, \(E_1=(1/2,1]\) and \(E_2=(1/4,1/2]\), so \(E_2\) is not a subset of \(E_1\). It would therefore be incorrect to apply continuity from above directly to \((E_n)\). The correct decreasing family is \(B_N=\bigcup_{n\ge N}E_n\). In that example \(m(E_n)=2^{-n}\), while \(m(B_N)=2^{-(N-1)}\); these are different quantities, and it is the tail unions whose decreasing intersection is controlled by almost-everywhere convergence.
The proof uses a useful general pattern: fix one error threshold, collect all later errors into a tail set, and then apply continuity from above to that decreasing family. On a finite measure space this converts pointwise eventual control outside a null set into convergence in measure. Finiteness is not a technical decoration; it is the condition that makes the tail-set measure argument work.
Check Your Understanding
For each question, keep the threshold convention \(|f_n-f|>\varepsilon\) and distinguish the error sets from their tail unions.
- In the powers example, find \(m(E_n(\varepsilon))\) when \(0<\varepsilon<1\), and explain why the point \(x=1\) does not prevent almost-everywhere convergence to zero.
- For a decreasing sequence of measurable sets \(A_N\) with \(\mu(A_1)<\infty\), identify the role of the finite-measure hypothesis in continuity from above.
- Explain why almost-everywhere convergence implies \(\bigcap_N\bigcup_{n\ge N}E_n(\varepsilon)\) is null for each fixed \(\varepsilon>0\).
- For the disjoint intervals \(I_n=(2^{-n},2^{-(n-1)}]\), compute the tail-union measure \(m(\bigcup_{n\ge N}I_n)\).
- In the countable-space example, which part of the finite-measure proof cannot be used, and what is \(\mu(E_n(\varepsilon))\) for \(0<\varepsilon<1\)?