I'd like to show that for an integrable sequence of functions $f_n:X \rightarrow [0, \infty)$ with $\sup_{n\geq 1} \int_{X} f_n d\mu < \infty, f_n \rightarrow f$ pointwise a.e. for some function $f:X \rightarrow [0,\infty)$ then the claim is that:
$\int_X(f_n) \rightarrow\int_X(f)$ iff $\int_X(|f_n-f|) \rightarrow 0$
I'd like to show the direction $"\Rightarrow"$ and was told that I would probably need Fatou's lemma and the inequality $|x – y| \leq |x| + |y|$, but I'm stuck. Could somebody please give me a hint?
What I've done:
I've said that $\int_X(f_n) \rightarrow\int_X(f) \Rightarrow \lim(\int_X f_n d\mu – \int_X f d\mu) =\lim (\int_X f_n -f d\mu) = 0$ and also that $\lim( \int |f_n -f| d\mu) \geq \int \lim(|f_n -f|) d\mu$ but I feel like this is the wrong direction for what I want to show.
Then I tried the approach $\lim \int |f_n – f| d\mu \leq \lim \int f_n d\mu+ \int f d\mu$ but didn't really know what to do with that 🙂
Any help would be greatly appreciated! Thanks!
Best Answer
If we're given a sequence of functions $f_n \geq 0$, $\mu$-integrable such that $f_n \to f$ pointwise almost everywhere, where $f$ is $\mu$-integrable as well, then $$\int_X f_n \, d\mu \to \int_X f \, d\mu \text{ if and only if } \int_X \vert f_n - f \vert \, d\mu \to 0.$$
For $(\Leftarrow)$, simply observe that if $\displaystyle \int_X \vert f_n - f \vert \, d\mu \to 0$, then for all $\varepsilon > 0$, there exits $N$ such that if $n > N$, then $\displaystyle \int_X \vert f_n - f \vert \, d\mu < \varepsilon$. This, together with the inequality: $$\bigg \vert \int_X f_n \, d\mu - \int_X f \, d\mu \bigg \vert = \bigg \vert \int_X (f_n - f) \, d\mu \bigg \vert \leq \int_X \vert f_n - f \vert \, d\mu$$ implies that $\displaystyle \int_X f_n \, d\mu \to \int_X f \, d\mu$ as desired.
For $(\Rightarrow)$ the converse implication, it is known as Scheffe's Lemma and this post has a few complete solutions.