Let {$f_n$} be a uniformly bounded sequence of functions which are Riemann integrable on $[0,1]$ and $$F_n (x) = \int_{0}^{x} f_n(t)dt \hspace{0.2in} ( 0 \leq x \leq 1 ) $$ I need to prove that there is a subsequence {$F_{n_k}$} which converges uniformly on $[0,1]$

I tried using the boundedness but got stuck right there !



By here, it suffices to show $\{F_n\}$ is pointwise bounded and equicontinuous.

Since $\{f_n\}$ is uniformly bounded, there exists some $M > 0$ such that $|f_n(t)| < M,\forall t \in [0,1]$ and all $n$. Therefore, $$|F_n(x)|=|\int_{0}^{x} f_n(t)dt|\leq \int_{0}^{x} |f_n(t)|dt \leq M(x−a) \forall n,$$ Hence we have proved pointwise boundedness of $\{F_n \}$.

For equicontinuity, given some $\epsilon > 0$, choose $\delta = \epsilon/M$ hence $\forall x, y \in [0,1]$ such that $|x − y| < \delta$, we have $$| F _n ( x ) − F_ n ( y ) | = |\int_{0}^{x} f_n(t)dt-\int_{0}^{y} f_n(t)dt|\leq \int_{y}^{x} |f_n(t)|dt\leq M|x-y|< \epsilon.$$ Hence we have done.


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