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Suppose that $f_n, f$ are non-negative measurable functions with $\mu(f_n)$ and $\mu (f)$ finite and such that $f_n\to f \text{ a.e.}$, Then $\mu(|f_n-f|)\to 0 \iff \mu(f_n)\to\mu(f)$.

For $\mu(f_n)\to\mu(f) \Rightarrow \mu(|f_n-f|)\to 0$

Using dominated convergence theorem, setting $g_n=f-f_n$ and first assuming that $\mu(f_n) \to \mu(f)$ and $|f_n-f|\to 0 \text{ a.e.}$

Then $\mu(|g_n|)=2\mu(g^+_n)-\mu(g_n)\to 0$ because $0\leq g~_n^+\leq f$ and $|f_n-f|\to 0\text{ a.e.}$

Why can you assume that $|f_n-f|\to 0 \text{ a.e.}$? Is it equivalent to $f_n\to f \text{ a.e.}$?

shilov
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1 Answers1

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Yes, $$ f_n\to f\;\text{ a.e.} \iff |f_n-f|\to\;\text{ a.e.} $$ This is simply because $$ f_n(x)\to f(x)\iff |f_n(x)-f(x)|\to 0,\quad \text{for any } x\in X, $$ so you can use the same null-set.

Stefan Hansen
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