Does $f_n\to0$ pointwise + $f_n$ integrable + $f_n$ uniformly bounded imply $\int f_n\to 0$? If $\{f_n:[0,1]\to\mathbb{R}\}_{n\ge 1}$ is a sequence of Riemann-integrable functions that converge pointwise to the zero function and $\{f_n\}_{n\ge 1}$ is uniformly bounded by $M$ can we prove that $\int_{[0,1]}f_n\to0$ ?
It seems difficult both to prove or disprove it. For a possible counterexample I came up with the sequence
$$f_1(x)=1-|2x-1|$$
$$f_{n+1}=1-2|f_n(x)-1/2|$$
which is the sequence of "spikes" at $x=j/2^n$ for $j=1,...,2^{n-1}$ so that $\int_{[0,1]}f_n=1/2$ for every $n\in\mathbb{N}$ but $f_n(x)$ it doesn't seem to converge to $0$ at some values like $x=1/3$. While we can get some $j/2^n$ arbitrarily close to every value, the closer we get, the crazier the spikes get.
For a possible proof, the set $A_{n,\varepsilon}:=\{\,x:|f_n(x)|\ge\varepsilon\,\}$ should in some sense be small so that we can separate the integral (or at least every lower/upper sum) of $f_n$ into the $A_{n,\varepsilon}$ with "large but bounded" $f_n$ but small interval width and the $[0,1]\setminus A_{n,\varepsilon}$ part with small $f_n$ and whatever interval. So let $P_m:=\{I_1,...,I_m\}$ be the partition of $[0,1]$ with $I_j=[\frac{j-1}{m},\frac{j}{m}]$. Now $J_1,...,J_r$ the $I_j$'s that contain points of $A_{n,\varepsilon}$ and $K_1,...,K_s$ the remaining $I_j$'s.
$$U(f_n,P_m)=\sum_{i=1}^rM_{J_i}\frac{1}{m}+\sum_{i=1}^sM_{K_i}\frac{1}{m}\le\frac{rK}{m}+\varepsilon$$
We at least know the limit $r/m$ exists as $m\to\infty$ and depends on $n$ because $f_n$ is integrable. Do you know of any proof or counterexample?
Thanks!

It is true, by Lebesgue's dominated convergence theorem. Just take $M>0$ such that$$(\forall n\in\mathbb N)\bigl(\forall x\in[0,1]\bigr):\bigl\lvert f_n(x)\bigr\rvert\leqslant M.$$Then, if you define $g(x)=M$ for each $x\in[0,1]$, $g$ is integrable and$$(\forall n\in\mathbb N)\bigl(\forall x\in[0,1]\bigr):\bigl\lvert f_n(x)\bigr\rvert\leqslant g(x).$$
