# 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$$(\forall n\in\mathbb N)\bigl(\forall x\in[0,1]\bigr):\bigl\lvert f_n(x)\bigr\rvert\leqslant g(x).$$