# Thread: [SOLVED] Integrability of an infinite series of functions

1. ## [SOLVED] Integrability of an infinite series of functions

I need to prove that if the sequence of functions $\displaystyle \{ f_n \}$
converges uniformly to $\displaystyle f$ on $\displaystyle [0,1]$, and each $\displaystyle f_n$ is integrable, then the limit function $\displaystyle f$ is also integrable on [0,1].

Does this help at all??
Since $\displaystyle \{ f_n \}$ converges uniformly to $\displaystyle f$, then
$\displaystyle \exists N$ such that $\displaystyle n>N \Rightarrow |f(x)-f_n(x)|<\varepsilon$

$\displaystyle \Rightarrow -\varepsilon<f(x)-f_n(x)<\varepsilon \Rightarrow f_n(x)-\varepsilon<f(x)<f_n(x)+\varepsilon$

2. Originally Posted by akolman
I need to prove that if the sequence of functions $\displaystyle \{ f_n \}$
converges uniformly to $\displaystyle f$ on $\displaystyle [0,1]$, and each $\displaystyle f_n$ is integrable, then the limit function $\displaystyle f$ is also integrable on [0,1].

Does this help at all??
Since $\displaystyle \{ f_n \}$ converges uniformly to $\displaystyle f$, then
$\displaystyle \exists N$ such that $\displaystyle n>N \Rightarrow |f(x)-f_n(x)|<\varepsilon$

$\displaystyle \Rightarrow -\varepsilon<f(x)-f_n(x)<\varepsilon \Rightarrow f_n(x)-\varepsilon<f(x)<f_n(x)+\varepsilon$
First let us remark that $\displaystyle \lim_{N\to\infty}S_N=\sum_{n=1}^{\infty}a_n$ where $\displaystyle S_N$ represents the $\displaystyle N$th partial sum. So now $\displaystyle \sum{a_n}\text{ converges}\leftrightarrow~S_N\text{ converges}$

1. So now we know that for a sequence $\displaystyle \left\{S_N\right\}$ to be convergent it must be montonic and bounded. Therefore $\displaystyle \exists{M}~\backepsilon\lim_{N\to\infty}S_N\leqsla nt{M}\stackrel{\text{in turn}}{\implies}\sum{a_n}\leqslant{M}$. So for every point where $\displaystyle \sum{f_n(x)}$ converges we have by neccessity $\displaystyle \exists{M}\backepsilon~\sum{f_n(x)}\leqslant{M}$. And if $\displaystyle \forall{x}\in[0,1]~\sum{f_n(x)}\stackrel{\text{uniformly}}\longright arrow{f}$ then $\displaystyle \forall{x}\in[0,1]\exists{M}\backepsilon~f\leqslant{M}$. Thus on $\displaystyle [0,1]$ $\displaystyle f$ is bounded.

2. Before we start the next step we are given that $\displaystyle f_n(x)$ is Riemann integrable, thus bounded and continous.

3.Next we just use the theorem that if $\displaystyle \forall{x}\in[a,b]~~\sum{f_n(x)}\stackrel{\text{uniformly}}\longrigh tarrow{f}$ and $\displaystyle f_n(x)$ is continuous on $\displaystyle [a,b]$ then $\displaystyle f$ is continous on $\displaystyle [a,b]$. So the fact that $\displaystyle \sum{f_n(x)}\stackrel{\text{uniformly}}\longrighta rrow{f}\wedge{f_n(x)}\in\mathcal{C}~~\forall{x}\in[0,1]$ tells us that $\displaystyle f$ is continuous on $\displaystyle [0,1]$

4. So we have shown that on $\displaystyle [0,1]$ $\displaystyle f$ is both bounded and continous. So it is Riemann integrable.