Let $M = \bigcup\limits_{n=1}^\infty M_n$, $M_n$ be a measurable sets on $\mathbb R$ and $f$ be a measurable function on $\mathbb R$. Then I want to know how the equality $$ \lim\limits_{N\to \infty}\int_{M\setminus \bigcup\limits_{n=1}^NM_n} |f(x)| \,dx = 0 $$ holds. It seems something like monotone convergence theorem, but I know the monotone convergence theorem is for the monotone "functions". How can I justify this?
Monotone convergence theorem for sets?
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$\begingroup$
real-analysis
integration
monotone-functions
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0$\int|f|<\infty$ is required. See my comment answering LeGrandDODOM. – 2017-01-09
2 Answers
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The map $\nu:A\mapsto \int_A |f|$ is a measure.
Note that $A_N:=M\setminus \bigcup_{n=1}^NM_n$ is a decreasing sequence that converges to $\emptyset$.
Assuming $\int |f|<\infty$, we have $\lim_N\nu(A_N)=\nu(\emptyset)=0$.
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Hint: use characteristic functions to write each integral as an integral on $M$.
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0Can you solve it without the assumption $\int |f|<\infty$ ? It seems to me the sequence of functions at hand is decreasing, so some domination is needed... – 2017-01-09
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0@LeGrandDODOM, without the assumption is obviously false. Take $f=1$, $M_n=[-n,n]$. – 2017-01-09