Lebesgue integral of continous function Let $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$ be a measure space. Let $g:\mathbb{R}\rightarrow \mathbb{R}$, be a continous function with
$$g_n(x)=g(x^n)\mathbb{1}_{[0,1]}(x), \qquad x\in \mathbb{R}, n\in \mathbb{N}.$$
I need to calculate
$$\lim_{n\rightarrow \infty} \int g_n d\lambda$$
And I am not sure if I have done this correct, since it depends on the function $g$, anyways, I have done the following:
\begin{align}
\lim_{n\rightarrow \infty} \int g_n d\lambda &=\lim_{n\rightarrow \infty} \int g(x^n)\cdot \mathbb{1}_{[0,1]}(x)\lambda(dx)\\
&=\lim_{n\rightarrow \infty} R\int_0 ^1g(x^n)dx \qquad (g:continous)\\
&=\lim_{n\rightarrow \infty}\left[g(x^n)x\right]_0^1=g(1).
\end{align}
Any help is much appreciated.
 A: Hint: 


*

*Compute the pointwise limit of $g_n$

*Use a theorem to change limes and integral.

A: Hint: What is the limit of $x^n$ as $n\to\infty$ and $x\in[0,1]$?
A: There is a problem in the passage $\lim_{n}\int_0^1{g(x^n)dx}= \lim_{n}xg(x^n)\big|_{0}^{1}$. This equality is generally false and I don't think this approach will help you.
The other answers suggest that you might be able to obtain, instead,
$\lim_{n}\int g_n(x)dx = \int \lim_{n}g_n(x)dx$.
The standard results for passing the limit into an integral are the
monotone convergence theorem and dominated convergence theorem.
For this problem, you can use the latter. You need to:


*

*Find an integrable function, $h(x)$, such that $|g_{n}(x)| \leq h(x)$.

*Find a function, $f(x)$ such that $f(x)=\lim_{n}g(x^n)$.


For (1), you can use the continuity of $g(x)$ and the fact that $g_{n}(x)$ is $0$ outside of $[0,1]$ to find $h(x)$. For (2), you can use the continuity of $g$ to obtain that, for every $x \in [0,1]$, $\lim_{n}g(x^n)=g(\lim_{n} x_n)$.
Once you verify these conditions, then you obtain
\begin{align*}
 \lim_n\int_0^1 g_n(x)dx
 &= \int_0^1 \lim_n g_n(x)dx
 = \int_0^1 f(x) dx
\end{align*}
