I am having some conceptual difficulties with almost everywhere (a.e.) convergence versus convergence in measure.
Let $f_{n} : X \to Y$. In my mind, a sequence of measurable functions $\{ f_{n} \}$ converges a.e. to the function $f$ if $f_{n} \to f$ everywhere on $X$ except for maybe on some set of measure zero.
However, the definition of convergence in measure is that $f_{n} \to f$ in measure if for every $\varepsilon>0$
\begin{equation}
\mu(\{x:|f_{n}(x)-f(x)|\geq\varepsilon\})\to0 \text{ as } n\to \infty
\end{equation}
Heuristically, it seems to me that convergence in measure is just saying that the subset of $X$ for which $f_{n}$ does not converge to $f$ must have a measure of zero.
I cannot see the difference between a.e. convergence and convergence in measure. If anyone can point out my conceptual error it would be much appreciated.
Best Answer
This is one of those things that is helpfully studied using an example. A very nice example for this issue is the "wandering block". Informally, the wandering block is the sequence of indicator functions of $[0,1],[0,1/2],[1/2,1],[0,1/4],[1/4,1/2],[1/2,3/4],[3/4,1]$, etc. More explicitly, it is the sequence $g_n(x)$ which comes about from enumerating the "triangular array" $f_{j,k}(x)=\chi_{[j2^{-k},(j+1)2^{-k}]}(x)$, where $k=0,1,\dots$ and $j=0,1,\dots,2^{k}-1$.
The sequence $g_n$ converges in measure to the zero function. You can see this as follows. Given $n$, write $g_n=f_{j,k}$, then $m(\{ x : |g_n(x)| \geq \varepsilon \})=2^{-k}$ for any given $\varepsilon \in (0,1)$. Since $k \to \infty$ as $n \to \infty$, this measure goes to $0$ as $n \to \infty$.
On the other hand, the sequence $g_n(x)$ does not converge at any individual point, because any given point is in infinitely many of these intervals and also not in infinitely many of these intervals. Thus the sequence $g_n(x)$ contains infinitely many $1$s and infinitely many $0$s, and so it cannot converge.
On an infinite measure space, there is an example for the other direction: $f_n(x)=\chi_{[n,n+1]}(x)$ on the line converges pointwise to $0$ but does not converge in measure, since $m(\{ x : |f_n(x)| \geq \varepsilon \})=1$ for $\varepsilon \in (0,1)$. A corollary of Egorov's theorem says that this is impossible on a finite measure space.
On a related note, the wandering block example also shows a nice, explicit example of the theorem "if $f_n$ converges in measure then a subsequence converges almost everywhere". Here, for any fixed $j$, the sequence $h_k=f_{j,k}$ (defined for sufficiently large $k$ that this makes sense) converges almost everywhere.