Uniformly bounded sequence of Riemann integrable functions Let {$f_n$} be a uniformly bounded sequence of Riemann int'ble functions on $[a,b]$.If $f_n\rightarrow 0$ pointwise then does it follow that $\int _{[a,b]}f_n\rightarrow0$?
My thoughts: The result doesn't follow from the given assumptions. To justify my claim, I choose $f_n(x)=\frac{x^2}{x^2+(1-nx)^2}$ on $[0,1]$ which satisfies all the criteria. Clearly, $f_n\rightarrow 0$ pointwise but I haven't been able to show that $\int _{[a,b]}f_n$ doesn't converge to $0$ although it's clear that it doesn't.
Are there any other counter-examples to justify this result?. I came up with $f_n(x)=nx(1-x^2)^n$ on $[0,1]$ but this choice of function doesn't have the uniform boundedness. Can anybody provide me with a relatively easy example to go with?
 A: Arzela's bounded convergence theorem (1885) states that if $(f_n)$ is a uniformly bounded sequence of Riemann integrable functions on $[a,b]$ that converges pointwise to a Riemann integrable function $f$, then $\int_a^b f_n(x) \, dx \to \int_a^b f(x) \, dx$.
In this case $f = 0$ is Riemann integrable and $\int_a^b f_n(x) \, dx \to 0$ follows. 
The assumptions are stronger than in convergence theorems for Lebesgue integrals since the Riemann integrability of the limit function is imposed.  This can be proved without measure theory using elementary techniques, for example here
.
A: As an immediate consequence of DCT and the fact that RI functions are also Lebesegue integrable (with the same value for the integral) we have $\int_{[a,b]} f_n(x) dx \to 0$.
A: Perhaps you can do it in this way, still measure-theoretic: By Egorov, given $\epsilon>0$, some measurable set $S\subseteq [0,1]$ is such that $f_{n}\rightarrow 0$ uniformly on $[0,1]-S$ and $|S|<\epsilon$, then $\left|\displaystyle\int f_{n}\right|=\left|\displaystyle\int_{[0,1]-S}f_{n}+\int_{S}f_{n}\right|\leq\left|\displaystyle\int_{[0,1]-S}f_{n}\right|+\sup_{n}|f_{n}(x)||S|\leq\left|\displaystyle\int_{[0,1]-S}f_{n}\right|+\left(\sup_{n}|f_{n}(x)|\right)\cdot\epsilon$. We know that $\displaystyle\int_{[0,1]-S}f_{n}\rightarrow 0$ by the uniform convergence of $f_{n}\rightarrow 0$ on $[0,1]-S$, so $\left|\displaystyle\int f_{n}\right|$ is arbitrarily small.
