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$(f_n), f$ are integrable. If $\int |f_n - f| \rightarrow 0$, then show that $\int_E f_n \rightarrow \int_E f$ for all measurable sets $E$ and that $\int f_n^+ \rightarrow \int f^+$.

Attempt: $$\left| \int f_n - \int f \,\right| \leq\int |f_n - f| \rightarrow 0 $$ which gives $\int f_n \rightarrow \int f$. How do I get this down to measurable sets $E$.

Any help is appreciated. Thanks

user62089
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1 Answers1

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Hint 1: $\int_E|g|\leq \int_X|g|$ where $X$ is your full measure space.

Hint 2: using the fact that $$y^+=\max(y,0)=\frac{|y|+y}{2}$$ we can see that $y\longmapsto y^+$ is $1$-Lipschitz: $$ |y^+-z^+|=\lvert\frac{|y|-|z|+y-z}{2} \rvert\leq\frac{||y|-|z||+|y-z|}{2}\leq |y-z|. $$

Julien
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