Show that $\int_{0}^{1}\frac{1}{1-f(t)}dt=\sum_{n=0}^{\infty}\int_{0}^{1}f(t)^n dt$.

Question

Let $f:[0,1]\rightarrow[0,1)$ be continuous. Then Show that $\int_{0}^{1}\frac{1}{1-f(t)}dt=\sum_{n=0}^{\infty}\int_{0}^{1}f(t)^n dt$.

What I know: $f$ is bounded by a constant function of $g=1$, which is integrable. Since $f$ is continuous on a compact set, it is uniformly continuous.

Question: (1) If instead of $\sum_{n=0}^{\infty}\int_{0}^{1}f(t)^n dt$, but $\sum_{n=0}^{n}\int_{0}^{1}f(t)^n dt$, can we movie the summation sign inside? Why? (2) If the answer is yes to (1), should $\sum_{n=0}^{\infty}f(t)^n dt=\int_{0}^{1}\frac{1}{1-f(t)}dt$?How?

Answer 1

Since $f(t)$ is continuous and $[0,1]$ is compact, it follows that the image of $[0,1]$ under $f$ is compact. Therefore there is some $c<1$ such that $0\leq f(t)\leq c$ for all $t$. Hence $$ \frac{1}{1-f(t)}=\sum_{n=0}^{\infty}f(t)^n$$ and moreover this holds uniformly in $t$ since $0\leq f\leq c<1$, so $$ \int_0^1\frac{1}{1-f(t)}\;dt=\int_0^1\sum_{n=0}^{\infty}f(t)^n\;dt=\sum_{n=0}^{\infty}\int_0^1f(t)^n\;dt$$

Answer 2

Because $f$ is non-negative, you have $\sum_{n=0}^\infty\int_0^1 f(t)^n\,dt =\int_0^1\left[\sum_{n=0}^\infty f(t)^n\right]\,dt$ by Tonelli's theorem. To finish use the familiar geometric series.