If a function is the limit in $L^1$ of simple functions, then that function is integrable

Question

I'm trying to prove that the completion of the integrable simple functions on $ (X,\Sigma,\mu) $ is $L^1(X,\Sigma,\mu)$. Suppose we have:

$lim_{n\to \infty}||f-s_n||_1=0$ for some sequence $(s_n)_n$ of simple integrable functions. How can I prove that $\int{|f(x)|}d\mu < \infty$?

score 1 · Accepted Answer · answered Mar 17 '21 at 16:12

1

Write $$\int |f| \ d\mu \leq \int |f - s_n| \ d\mu + \int |s_n| \ d\mu,$$ where $n$ is taken to be sufficiently large.

answered Mar 17 '21 at 16:12

Gary Moon

How can you know the second term converges to zero? – Othman El Hammouchi Mar 17 '21 at 16:21
@othi It doesn't converge to $0$, but it is finite, which is what you need to show that $f \in L^1(d\mu)$. – Gary Moon Mar 17 '21 at 16:22
But then why do you have to take n sufficiently large? I don't get it... – Othman El Hammouchi Mar 17 '21 at 16:25
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@othi You take $n$ sufficiently large to get some kind of bound, really any bound, on the first term. You could take $n$ sufficiently large to make the first term less than $1$, say. Then, you have $\int f \ d\mu < \int |s_n| \ d\mu + 1 < +\infty$. – Gary Moon Mar 17 '21 at 16:27
Thanks a lot!!! – Othman El Hammouchi Mar 17 '21 at 16:29
@othi My pleasure. – Gary Moon Mar 17 '21 at 16:30

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