Limiting behavior of sequence of integrals Let $f$ and $g$ be continuous on $I : \{a\le x \le b\}$ with $f(x)\ge 0$ and $g(x)>0$.  Let $M= max_I f(x)$ and the sequence $\{M_k\}$ be defined by
$$M_k=\int_a^b g(x)[f(x)]^k dx$$
The problem is to show that $M_{k+1}/M_k \rightarrow M$ if $M > 0$.
It is easy to see that  $M_{k+1}/M_k \le M$ since $f(x) \le M$ implies $$\int_a^b g(x)[f(x)]^{k+1} dx \le M\int_a^b g(x)[f(x)]^k dx$$
The sequence can also be written as
$$M_k=M^k\int_a^b g(x)\left[ \frac{f(x)}M \right]^k dx$$
In the limit, the integrand becomes discontinuous ($g(x)$ if $f(x) = M$ and $0$ elsewhere), so theorems involving uniform convergence don't apply.  I'm not sure how to proceed.
 A: Noting $f_n(x) = g(x) \left( \frac{f(x)}{M} \right)^n$ we see that the dominated convergence theorem hypotheses are fulfilled and that $\int_a^b f_n(x)dx$ converges to $\int_a^b g(x)\mathbf{1}_{\{t\in [a,b]\;|\;f(t)=M\}}(x)dx$. Then as $M_k=M^k\int_a^b g(x)\left[ \frac{f(x)}M \right]^k dx$ your see that $M_k$ is equivalent to $I M^k$, so that $M^{k+1} / M^k$ converges indeed to $M$ if $I > 0$. Now, do you see how to deal with the case where $I = 0$ ? By the way, do you see if we really need to distinguish cases $I >0$ and $I = 0$ ?
Remark. Here we used the dominated convergence theorem for the Lebesgues integral, as as you remarked indeed, the limit function may not be continuous. But could we use the same theorem for the Riemann integral ? To rephrase the question : are functions $x\mapsto \mathbf{1}_{\{t\in [a,b]\;|\;f(t)=c\}}(x)$ Riemann integrable as soon as $f$ is Riemann integrable ?
A: So I now have the following.  Let
$$S_k=\int_a^b g(x)\left[ \frac{f(x)}M \right]^k dx$$
$S_k$ is a nonincreasing sequence bounded below by $0$.  Hence $S$ = lim $S_k$ exists.  Since $M_k=M^kS_k$, 
$$lim \frac{M_{k+1}}{M_k} = lim \frac{M^{k+1}S_{k+1}}{M^kS_{k}} = M$$
provided $S \ne 0$.  Now I have to consider the case when $S = 0$.
