If the measure of X is infinity and fn converges to 0 a.e. then is it true that fn converges to 0 in measure?

Question

I was thinking of the example in Folland on page 61 i.e. $\mu(\mathbb{R})=\infty$. Let $f_n=n\chi_{[0,1/n]}\rightarrow 0$ a.e. Then $f_n\rightarrow 0$ in measure. My inclination is that this is true so it requires proof. Also the same question is asked but for when $\mu(X)=1$. My inclination is that this is not true and a counterexample can be provided.

Answer 1

The statement that convergence a.e. implies convergence in measure is true in finite measure spaces, not true in general.

For the first statement, note $f_n\to f$ a.e. implies for all $\varepsilon>0$ and $x\in X$, there exists $n$ such that $k\geq n$ implies $|f_n(x)-f(x)|<\varepsilon$. In particular, this implies $$\bigcap_n\bigcup_{k\geq n}\{x\in X:|f_k(x)-f(x)|\geq\varepsilon\}=\emptyset.$$ Finiteness of the measure space gives lets us use continuity from above to yield $$\lim_n \mu\left(\bigcup_{k\geq n}\{x\in X:|f_k(x)-f(x)|\geq\varepsilon\}\right)=0$$ which implies convergence in measure.

To see how this fails in general measure spaces, consider $f_n=\mathbf{1}_{(-\infty,n]}$, which converges to $1$ everywhere but not in measure.