Almost Everywhere Convergence versus Convergence in Measure 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. 
 A: On finite measure spaces, almost sure convergence is a stronger property than convergence in measure. A sequence converges almost surely if there is a fixed nullset $N$ so that $f_n(x)$ converges to $f(x)$ for all $x \notin N$.
However, we don't need such a set for the convergence in measure. We just say that the set on which $f_n$ and $f$ differ by more than $\epsilon$ should become arbitrary small if $n \to \infty$.
Maybe it will help your understanding to see an example of a sequence that converges in measure, but not almost surely. I've posted such a sequence here.
A: 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.
