# If $f_n$ converges to $f$ in measure then $f$ is finite almost everywhere

Let $$A$$ be a measurable set with finite measure, $$f_n$$ converges to $$f$$ in measure on $$A$$. May I conclude that $$f$$ is finite almost everywhere on $$A$$?

Personally, I think the answer is "yes". By contradiction, $$f$$ is not almost everywhere. It is implied that there exists a set $$B$$ with positive measure such that $$f$$ is infinite on $$B$$. Thus, $$\vert f_n - f \vert$$ is infinite on $$B$$, for all $$n \in \mathbb{N}$$. Therefore, for all $$\varepsilon > 0$$ $$\{x \in A : \vert f_n - f \vert \ge \varepsilon \} \supset B .$$ So, $$\mu(\{ x \in A : \vert f_n - f \vert \ge \varepsilon \}) >0.$$ This implies that $$\lim\limits_{n \to \infty} \mu(\{ x \in A : \vert f_n - f \vert \ge \varepsilon \}) > 0$$. Therefore, $$f_n$$ doesn't converge to $$f$$ in measure on $$A$$ which is clearly absurd.

I'm not sure about my conclusion. Please give me some hints regard this.

The conclusion is false. First observation, a positive sequence can converge to zero ( e.g. $$(1/n)$$ ). Apart from this, you have essentially proved that for $$(f_n)$$ to converge in measure to $$f$$, the positive measure set where $$f$$ is infinite must be the same (up to a negligible set) as the one in $$(f_n)$$ for all $$n$$ from a certain rank. But there is nothing from this preventing $$f$$ from having a positive measure set where it is infinite. (A trivial example would be the constant sequence $$f_n =f$$).
• Why would that be ? All I am saying is that we do not need $f$ to be bounded almost everywhere. For example, let be $f:\mathbb{N} \to \mathbb{R} \cup [-\infty,\infty]$ with $f(0) = \infty$ and $f(n) = 0$ otherwise and $\mu$ the coutning measure. Then, for $(f_n)$ to converge in measure to $f$, it is necessary that $f_n(0) = \infty$ for all $n \ge N$ for some $N$. – nicomezi May 17 at 9:35