1. ## continuous, sequence, integral

Suppose that $\displaystyle f: [0,1] \rightarrow \mathbb{R}$ is continuous. For each $\displaystyle n \in \mathbb{N}$, define $\displaystyle g_n : [0,1] \rightarrow \mathbb{R}$ by $\displaystyle g_n(x)=f(x^n)$ for all $\displaystyle x \in [0,1]$. Prove that the sequence $\displaystyle (\int^1_0 g_n(x))$ converges to $\displaystyle f(0).$
Attempt
This question really confuses me. I don't see how this sequence converges to $\displaystyle f(0)$. A few helpful hints on this one would be appreciated. Thanks in advance.

2. Hello,

Firstly, you have to prove that $\displaystyle \lim_{n \to \infty} \int_0^1 g_n(x) ~dx=\int_0^1 \lim_{n \to \infty} g_n(x) ~dx$, by using the dominated convergence theorem for example (f is continuous over a closed set, --> it is bounded)

Secondly, you have to prove that $\displaystyle g_n(x) \xrightarrow[n \to \infty]{} f(0)$
In order to show that, write the epsilon definition of this limit : $\displaystyle x^n \xrightarrow[n \to \infty]{} 0$, when $\displaystyle x \in [0,1)$ (you can "miss" the case x=1, because it's a neglectible case when dealing with integrals... dunno if you know how to deal with it)
Then use the epsilon-delta method on the continuity of f, and you're done.

3. Originally Posted by Moo
Hello,

Firstly, you have to prove that $\displaystyle \lim_{n \to \infty} \int_0^1 g_n(x) ~dx=\int_0^1 \lim_{n \to \infty} g_n(x) ~dx$, by using the dominated convergence theorem for example (f is continuous over a closed set, --> it is bounded)

Secondly, you have to prove that $\displaystyle g_n(x) \xrightarrow[n \to \infty]{} f(0)$
In order to show that, write the epsilon definition of this limit : $\displaystyle x^n \xrightarrow[n \to \infty]{} 0$, when $\displaystyle x \in [0,1)$ (you can "miss" the case x=1, because it's a neglectible case when dealing with integrals... dunno if you know how to deal with it)
Then use the epsilon-delta method on the continuity of f, and you're done.
What happens for the case $\displaystyle x=1$? Can we overlook the case $\displaystyle x=1$ or does this case also satisfy the requirement that $\displaystyle (\int^1_0 g_n(x))$ converges to $\displaystyle f(0)?$

4. Originally Posted by maursvolta41
What happens for the case $\displaystyle x=1$? Can we overlook the case $\displaystyle x=1$ or does this case also satisfy the requirement that $\displaystyle (\int^1_0 g_n(x))$ converges to $\displaystyle f(0)?$
It does satisfy the requirement. Because when dealing with an integral, a single point can be omitted (in the case of a continuous function)

If you know some measure theory, you can say that {1} is a set of Lebesgue measure equal to 0. So x^n converges almost everywhere to 0.