Convergence in measure There is a proposition that states the following: Assume $E$ has finite measure. Let {$f_n$} be a sequence of measurable functions that converges pointwise a.e. on $E$ to $f$ and $f$ is finite a.e. on $E$. Then {$f_n$} converges in measure to $f$ on $E$. How do you show that this fails if $E$ has infinite measure? 
 A: To see that the proposition is not true in general if $E$ does not have finite measure, you just need a counterexample.  My suggestion is to start with $E=\mathbb{R}$, and come up with a sequence of measurable functions $(f_n)_{n\geq 1}$ converging pointwise to $0$, while the measure of the set where $f_n(x)\geq1$ is always infinite.  For instance, if you have $f_n(x)=0$ when $x<n$, then the sequence will converge to $0$ at each point regardless of what $f_n(x)$ is when $x\geq n$.  This gives a lot of freedom for strange counterexamples.
A: We make the comment of Sachin clearer.
Choose $E=\mathbb{R}$ and $\mu$ is the Lebesgue measure on $E$. Consider the sequence of measurable functions $\{f_n(x)\}_{n\in \mathbb{N}}$ given by
$$
f_n(x)=\begin{cases}
1 & n\leq x\leq n+1, \\
0& \text{ortherwise}.
\end{cases}
$$ 
for all $x\in E$ and $n\in \mathbb{N}$.
Then:


*

*For every $x\in E$ we have $x\notin [n,n+1]$ or $f_n(x)=0$ for sufficiently large $n$ 
and so $$|f_n(x)-0|=|0-0|=0$$ for sufficiently large $n$. This implies that
$$\displaystyle\lim_{n\rightarrow\infty}f_n(x)= 0$$ for all $x\in E$.
Therefore the sequence $\{f_n(x)\}_{n\in \mathbb{N}}$ is pointwise convergent
to $f=0$.

*We observe that
$$
\left\{x\in E: |f_n(x)-0|\geq \frac{1}{2}\right\}=[n, n+1].
$$
and so
$$
\mu\left\{x\in E: |f_n(x)-0|\geq \frac{1}{2}\right\}=\mu([n, n+1])=1.
$$
Hence $\{f_n(x)\}_{n\in\mathbb{N}}$ is not convergent in measure to $f=0$.
