Assume that you have a measurespace $(A,\mathcal{A},\mu)$. And you sequence of measurable functions $f_n \rightarrow \mathbb{R}$, that are increasing, and each function is bounded below by a common value $-M$.

Do we then have that $\lim\limits_{n \rightarrow \infty}\int\limits_{A}f_n(x)d\mu=\int\limits_{A}\lim\limits_{n \rightarrow \infty}f_nn(x)d\mu$?

I am able to prove this for a finite measure space by considering the non-negative and increasing sequence $\{f_n+M\}$ and using the monotone convergint theorem. But does it hold for a measure-space with infinite measure?

The reason I don't get it to work with a general measure space is that the integral of the constant function $M$ may not be finite, so I get in a situation where I can't cancel the parts.


Not true. On the real line with Lebesgue measure let $f_n(x)=-1$ for $x \geq n$ and $0$ for $x <n$. Then $f_n \geq -1, (f_n)$ is increasing and $\lim \int f_n=-\infty \neq 0 =\int \lim f_n$

  • 1
    $\begingroup$ Thank you very much! $\endgroup$
    – user394334
    Jul 18 '20 at 0:15

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