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?
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.