Given f continues function in [0,1] we define a sequence of functions fn(x) = f(x^n) , given that for every number 0<a<1 the sequence of function is uniformly convergs in [0,a] to f(0) prove that the integral of fn(x) from 0 to 1 is f(0).
So i have tried this : Since the sequence of function uniformly converges to f(0) in [0,a] for every 0<a<1 , fn(x)uniformly converges in [0,1) , i defined a new sequence of functions Gn(x) = fn(x) for every x in [0,1) , and for x= 1 Gn(1)= f(0) . So Gn(x) uniformly converges to f(0) , now we can do the limit first then the integral , since the limit equals f(0) , we have now integral of f(0) from 0 to 1 which equals f(0) .
I want to know why my proof is wrong .
Let $\varepsilon>0$. Then there exists a $\delta>0$, such that $$ 0\le x <\delta\quad\Longrightarrow\quad |f(x)-f(0)|<\varepsilon. $$ If $a\in (0,1),\,$ then $a^n\to 0$, and hence there is an $n_0\in\mathbb N$, such that $$ n\ge n_0 \quad\Longrightarrow\quad 0<a^n<\delta. $$ Hence, if $n\ge n_0$ and $x\in [0,a],\,$ then $0\le x^n\le a^n<\delta,$ and $$ |f_n(x)-f(0)|=|f(x^n)-f(0)|<\varepsilon. $$ Indeed, $f_n|_{[0,a]}\to f(0)$ uniformly.
Next, $\varepsilon>0$, and let $M=\max_{x\in [0,1]} |f(x)|$, and $a\in (0,1),$ such that $M(1-a)<\varepsilon$. Then $$ \left|\int_0^1 f_n(x)\,dx-f(0)\right|\le \int_0^1 |f_n(x)-f(0)|\,dx\\=\int_0^a |f_n(x)-f(0)|\,dx+\int_a^1 |f_n(x)-f(0)|\,dx<a\varepsilon+2M(1-a)<(a+1)\varepsilon $$ for $n\ge n_0$.