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Fix a measure space $(X,\mathcal{M},\mu)$. Suppose $\{f_n\} \subset L^1$ and $f_n \to f$ uniformly. Then I want to prove the following statement:

If $\mu(X)<\infty$ then $f \in L^1(\mu)$ and $\int f_n \to \int f$.

In some lecture notes, the attempted solution is as follows (I rewrite it in my language):

Since $f_n \to f$ uniformly then there exists $N$ such that $\left\lvert f_n(x) - f(x)\right\rvert < 1$ for all $n \geq N$ and all $x \in X$. Therefore, $$\int \left\lvert f \right\rvert \leq \int \left\lvert f - f_N \right\rvert + \int \left\lvert f_N \right\rvert \leq \int 1+ \int \left\lvert f_N \right\rvert = \mu(X) + \int \left\lvert f_N \right\rvert < \infty$$

so $f \in L^1$.

Then main point that I do not understand here is that, how can we say that $\mu(X) + \int \left\lvert f_N \right\rvert < \infty$? $\mu(X) < \infty$ by definition, however I couldn't see why $\int \left\lvert f_N \right\rvert < \infty$. Is this solution makes sense? If so, can anyone explain this? Thanks!

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$f_N$ is integrable, isn't it? –  Davide Giraudo Jan 11 '13 at 21:22
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$f_N\in L^1(\mu)$, isn't it? –  Ilya Jan 11 '13 at 21:22
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@DavideGiraudo: it's even funny, that our comments with 8 seconds difference are equivalent, and still different :) –  Ilya Jan 11 '13 at 21:23
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@John: since you are done so fast, what if $\mu(X) = \infty$? –  Ilya Jan 11 '13 at 21:34
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Then I guess these conditions fail. I will try to construct counterexamples. –  Deniz Jan 11 '13 at 21:54

1 Answer 1

$$\left|\int_X f_n\,\mathrm d\mu-\int_X f\,\mathrm d\mu\right|\leqslant\int_X|f_n-f|\,\mathrm d\mu\leqslant\mu(X)\cdot\|f-f_n\|_\infty$$

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