TITLE: Showing $\sum_{n = 0}^\infty \int f^n$ converges
QUESTION [9 upvotes]: I am having trouble solving a real analysis qualifying exam problem.  

The question assumes $\mu(X) < \infty$ and $\left| f \right| < 1$ (EDIT: Assume $f$ is real-valued).  We are to show that $$ \lim_{n \to \infty} \int_X 1 + f + \dots + f^n d\mu$$ exists, possibly equal to $\infty$.  

My work so far.  Each integral in the sequence makes sense since $\int 1 + \left| f \right| + \dots + \left| f \right|^n < (n+1) \mu(X) < \infty$.  Rephrasing the problem, we want to show $\sum_{n = 0}^\infty \int f^n$ converges.  It is immediate by the Monotone Convergence Theorem that the result is true for nonnegative functions $f$.  Considering absolute convergence, we have $$\sum \left| \int f^n \right| \leq \sum \int \left| f \right|^n$$ where the series on the right converges by what we just said.  If said series is finite, then $\sum \int f^n$ converges absolutely, hence converges.  
Question.  I am stuck on the case that $$ \sum \int \left| f \right|^n = \infty. \;\;\;\;\;\;\;\;\; (*)$$    
I know from the statement of the problem that we are allowing for $\sum \int f^n = \infty$, but it is not clear to me whether this should follow from $(*)$.  We do know $$\sum \int \left| f \right|^n = \int \sum \left| f \right|^n = \frac{1}{1 - \left| f \right|}.$$  So if this equals $\infty$, then $\mu \left\{ x \colon \left| f(x) \right| > 1 - \frac{1}{n} \right\} > 0$ for all $n$.  And of course $\sum f^n = \frac{1}{1 - f}$ as well.  But I can't see how to put this all together.  
Any help would be much appreciated.  Thanks.

REPLY [2 votes]: For $f \geq 0$, you showed the claim yourself.  Now, for the general case, by splitting $f = f_+ - f_-$, it suffices to show the claim for $f\leq 0$. For this, also note $f^n = (f_+)^n + (-f_-)^n $, since the supports of $f_+, f_-$ are disjoint. 
Now, for $-1 < x\leq 0$, we have
$$
\bigg | \sum_{k=0 }^n x^k \bigg | = \frac {1-x^{n+1}}{1-x} \leq 1,
$$
so that you can apply the dominated convergence theorem. This easily shows that
$$
\lim_n \int 1+\dots + f^n d\mu = \int \frac {1}{1-f}d\mu 
$$
is finite (remember  $f \leq 0$). Together with the case $f\geq 0$, we get the claim.