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Is there any counter example for for $F_n \downarrow F$ then $\int F_n \, d\mu \downarrow \int F \, d\mu$?

I came up with the one below but $F_n$ does not go down to $0$ monotonically. I need something that goes monotonically.

$$F_n = \frac1n \cdot 1_{[0,n]}(x)$$

we know that $\int F_n \, d\mu = 1 \text{ and not $\int0 \, d\mu$}$

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    What is the relationship between $F_{n}$ and $f_{n}$? And by the way, in your example $F_{n}$ does not decrease. Note that for any $n\in\mathbb{N}$ we have $F_{n+1}(n+1)=\frac{1}{n+1}>0=F_{n}(n+1)$.2012-12-06
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    Hi Thomas, I fixed the issue.2012-12-06
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    Also say whether you know the monotone convergence theorem, and why it does not answer the question.2012-12-06
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    Hi @GEdgar, what a great world. Here is like a class for me that everyone are trying to help others to learn. Thanks a lot for your comment. Yes, I know monotone convergence theorem. In the theorem $f_n$ converges increasingly to $f$, but here is the other way.2012-12-06
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    The fact that $(f_n)$ is non-increasing is not a big problem in itself since you could consider $-f_n$ and $-f$. The reason why the monotone convergence theorem doesn't apply here is that it needs $-f_n$ to be **non-negative**, i.e. $f_n$ non-positive.2012-12-06

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Why not $f_n = 1_{[n,\infty)}$ and $f = 0$ ?

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    Thanks Julien. Could you say with what measure...Lebesgue measure?2012-12-06
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    Yes, for example.2012-12-06