This question is an extension of a question asked earlier.

Let $(X,\mathcal{M},\mu)$ be a measure space and let $f_{n}: X \to Y$, where $\{f_{n}\}$ is a sequence of functions. The proof wiki definition of Almost Everywhere (a.e.) Convergence is that $f_{n}$ converges almost everywhere on $X$ iff:

\begin{equation} \mu(\{ x \in X: f_{n}(x) \text{ does not converge to } f(x) \})=0 \end{equation}

It seems to me (and correct me if I am wrong) that we could easily rewrite this as $f_{n}$ converges almost everywhere on $X$ iff:

\begin{equation} \forall \varepsilon>0: \lim_{n\to \infty}\mu(\{ x \in X: |f_{n}(x) - f(x)|\geq \varepsilon \})=0 \end{equation}

Now proof wiki also gives the definition of Convergence in Measure, where it says $f_{n}$ converges in measure to $f$ iff:

\begin{equation} \forall \varepsilon>0: \lim_{n\to \infty}\mu(\{ x \in D: |f_{n}(x) - f(x)|\geq \varepsilon \})=0 \text{ for all } D\in \mathcal{M} \text{ with } \mu(D)<\infty \end{equation}

I have two questions:

  1. Is my re-characterization of the definition of a.e. convergence valid?

  2. Is the only difference between a.e. convergence and convergence in measure the fact that the limit in a.e. convergence is only required to hold on the entire space $X$ (infinite or not) whereas the limit is required to hold in convergence in measure for every finite set in the $\sigma$-algebra on $X$?

I suspect the answer to both these questions are "yes" but am looking for confirmation. Unlike previous questions on the topic I am not looking for examples, as I have not found them very helpful. If the answer to any of my questions is "no" I am looking for an answer that appeals directly to the definition of a.e. convergence or convergence in measure.


This is not correct. $f_n\to f$ a.e. iff $$\mu(\{x :\lim_{n\to\infty} |f_n(x)-f(x)|\ne0\}) = 0. $$ Convergence in measure has the limit outside the measure - one must take care when interchanging limit operations!

Now, convergence in measure implies that there is a subsequence that converges a.e., and in a finite measure space (i.e. $\mu(X)<\infty$), convergence a.e. implies convergence in measure. But this is all we can say in general, without further assumptions.

  • $\begingroup$ Is it then true that $f_{n}\to f$ a.e. on $X$ iff $\forall \varepsilon>0: \mu(\{ x \in X: \lim_{n\to \infty} |f_{n}(x) - f(x)| \geq \varepsilon \})=0$? $\endgroup$ – möbius Sep 1 '15 at 18:27
  • 3
    $\begingroup$ @möbius no, it is not true. $\endgroup$ – uniquesolution Sep 1 '15 at 18:33

This is not exactly an answer to the question as formulated, but it may clarify the precise formulation of the distinction. Convergence a.e. can be written using countable intersections and unions. First define

$$A_{n,m} = \{ x : |f_n(x) - f(x)| > 1/m \}.$$

Now, for the convergence to fail at $x$, there must be some $m$ such that $x \in A_{n,m}$ for infinitely many $n$. In general, the set $B_m$ defined by "$x \in B_m$ if and only if there are infinitely many $n$ such that $x \in A_{n,m}$" is given by

$$B_m = \bigcap_{k=1}^\infty \bigcup_{n=k}^\infty A_{n,m}.$$

Now if there must be some such $m$, then the bad set where convergence fails is

$$C=\bigcup_{m=1}^\infty \bigcap_{k=1}^\infty \bigcup_{n=k}^\infty A_{n,m}.$$

Convergence a.e. says that this set has measure zero. It is equivalent to say that

$$\mu \left ( \bigcap_{k=1}^\infty \bigcup_{n=k}^\infty A_{n,m} \right )$$

is zero for every $m$.


Yes, your re-characterization is wrong, becuase it is very well known that convergence in measure does not imply convergence almost everywhere. Here is an example. Consider the intervals $E_{i,n}=[{i-1\over n},{i\over n}]$ and their characteristic functions $\chi_{E_{i,n}}$, where for each $n$ $i$ runs between $1$ and $n$. So the few first intervals are $$E_{1,1}=[0,1], E_{1,2}=[0,{1\over 2}], E_{2,2}=[{1\over 2},1], E_{1,3}=[0,{1\over 3}], E_{2,3}=[{1\over 3},{2\over 3}]$$ and so on. Now, for every $\varepsilon>0$, $$m\{x\in[0,1] : \chi_{E_{i,n}}(x)>\varepsilon\}=m\{x\in [0,1]: \chi_{E_{i,n}}(x)=1\}={1\over n}\longrightarrow 0$$ Where $m$ denotes the Lebesgue measure. Thus the sequence $\chi_{E_{i,n}}$ tends to zero in the Lebesgue measure, but it does not tend to zero almost everywhere, because for every $x\in [0,1]$ there are infinitely many intervals $E_{i,n}$ that contain $x$, hence there will be a subsequence for which $\chi_{E_k}(x)=1$. So this is an example where a sequence converges in measure, but fails completely to converge point-wise, not even at a single point. I know you said you didn't want examples. But if you state a false mathematical statement, there is little use in a philosophical discussion as to why it is false, what are the deep reasons and so on, because frankly, there are no deep reasons. For example, if I were to state that every continuous function is differentiable, would you opt for an example or a discussion?


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