Help computing $\lim_{n\to \infty}\int_{0}^{1}f_{n}(x)\,dx$ We have the function $f(x)=2x(1-x)$ and we define $f^{\circ 2}=f\circ f$, and $f^{\circ n}=f\circ f^{\circ(n-1)}$ for $n>2$. We need to compute $$\lim_{n\to\infty}\int_{0}^{1}f_{n}(x)dx$$. 
To do that, I found that $f^{\circ n}(x)\leq \frac{1}{2}$ for all $x\in[0,1]$ and for all $n$. Furthermore, I found that $$f(x)\leq f^{\circ 2}(x)\leq f^{\circ 3}(x)\leq \cdots \leq f^{\circ n}(x)\leq \frac{1}{2}.$$
With this in mind I obtained that $$\lim_{n\to \infty }\int_{0}^{1}f^{\circ n}(x)\,dx\leq \frac{1}{2}$$. In addition, I guess that the sequence of functions $\{f^{\circ n}(x)\}_{n\in \mathbf{N}}$ converges pointwise to the function $g(x)=1/2$ for $x\neq 0,1$ and $0$ for $x=0,1,$ but I couldn't prove it.
Please any suggestion for this problem would be really appreciated. 

Your intuition regarding the limit of $(f^{\circ n}(x_0))$ for all $x_0 \in [0,1]$ is correct.
To prove it, you can prove that:


*

*The only possible limits of the sequence for $x_0 \in [0,1]$ are $0$ and $1/2$, i.e. the roots of the equation $f(x)=x$ as $f$ is a continuous map.

*For $x_0 \in (0,1/2)$, the sequence is increasing, bounded by $1/2$ and therefore converging to $1/2$.

*For $x_0 \in (1/2,1)$, you have $x_1 \in (0,1/2)$ and you can therefore use previous case.

