If $f_n$ is a sequence of integrable functions that converges uniformly, is it's limit integrable?

Question

Well, i want to prove that if $f_{n}$ is sequence of integrable functions on $[a,b]$, and $f_{n}\rightarrow f$ uniformly, then $f$ is also integrable on $[a,b]$.

So far, i divided the proof for two: for sequence of continuous functions, and for sequence of non-continuous functions.
The first one is i succeed to prove, but i don't know what to do for the second series...
Can someone help me? tnx!

Answer 1

the second one doesn't hold assuming you're talking about $\mathbb{R}$ with usual measure. take $f$ to be a function s.t. it's $0$ for negative $x$ and $\frac{1}{n}$ on the intervals $[n-1, n]$ for positive integer $n$. then $f$ is not integrable because $\sum \frac{1}{n}$ is divergent, while taking $f_n$'s to be $f$ 'cut out' after $n$, so $f_n(x) = f(x)$ for $x \leq n$ and $f_n(x) = 0$ for $x > n$ yields uniform convergence adn all $f_n$'s are integrable

also you can make this counterexample work for the continuous case as well by slightly regularizing your functions, so perhaps you're missing some assumptions?

edit: for the edited version of your question you just take $n$ large enough so that $|f_n - f| < 1$ for all $x$ and then you use $|f| \leq |f_n| + |f_n - f|$ - it's triangle inequality

Answer 2

For finite measure space, the uniform metric bounded above the $||\cdot||_{L_{1}}$ norm metric (ignoring a constant factor). Since $f_{n}$ converge uniformly, it is Cauchy in uniform metric, and thus is also Cauchy in $||\cdot||_{L_{1}}$ norm metric. Thus $f$ is integrable, as $L_{1}$ is a complete space.

Answer 3

If $\{f_n\}$ is a sequence of Riemann integrable functions, then the discontinuity set of each $f_n$ has Lebesgue measure zero. If $f$ is the uniform limit of $\{f_n\}$ and $f$ is discontinuous at $x$, then it must be that some $f_n$ is discontinuous at $x$. This is a corollary of the fact that if ${f_n}$ is a sequence of functions that are continuous at $x$ and they uniformly approach an $f$, then $f$ is continuous at $x$. Thus the discontinuity set of $f$ is contained in the union of the discontinuity set of the $f_n$s. Since this set is contained in a countable union of measure zero sets, it also has measure zero. Thus $f$'s discontinuity set also has measure zero and thus is Riemann integrable.