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Suppose $ f_n \subset L^1(\mu)$ and $f_n$ uniformly convergent to $f$. Could we say $\int f_n \to \int f$? Or the condition $\mu(X) < \infty$ must be needed?

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It's not true in general that if $f_n\to f$ uniformly then $\int f_n\to\int f$.

For instance, suppose that $X=\mathbb{R}$ with Lebesgue measure, and let $f_n(x)=\frac{1}{n}\cdot 1_{[0,n]}(x)$ and $f=0$. Then $f_n\to f$ uniformly on $\mathbb{R}$, but $\int f_n=1$ for all $n$.

However, if $\mu(X)<\infty$ then $$ \Big|\int f_n-\int f\Big|\leq \int |f_n-f|\leq \mu(X)\sup_{x}|f_n(x)-f(x)|\to 0$$ as $n\to\infty$.

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    I do not follow the counterexample. – 2017-02-22
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    Which part of it is confusing? – 2017-02-22
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    Definition of $f_n$. – 2017-02-22
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    $1_{[0,n]}(x)$ is the indicator function of the interval $[0,n]$. So $f_n(x)=\frac{1}{n}$ if $0\leq x\leq n$, and $f_n(x)=0$ otherwise. – 2017-02-22
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    If ${f_n}$ be monotone sequence of function then we have $\int f_n \to \int f$? – 2017-02-22
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    Yes, if the sequence $\{f_n\}$ is non-negative and increasing then $\int f_n\to\int f$ by the monotone convergence theorem. – 2017-02-22
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    If $f$ isn't measurable then it doesn't make sense to consider its integral. – 2017-02-22
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    I am sorry for having taken your time if $f\subset L^1$ what we can say? – 2017-02-22
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    Well, in my counterexample $f$ is in $L^1$, right? – 2017-02-22
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    Definitely. for conclusion if we have monotone condition (for decreasing $\int f < \infty$) it is true. Even we do not need uniformly convergent. _a.e._ is enough. Right? – 2017-02-22