# Increasing convergence of sequence bounded below.

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 $$x . Then $$f_n \geq -1, (f_n)$$ is increasing and $$\lim \int f_n=-\infty \neq 0 =\int \lim f_n$$