difficult problem in riemman integrals

Thanks to Motyla Noga Tomka Mazura for part of what's below - this is easier than I've thought for years, because of the monotonicity that he pointed out. (I think this post is about half mine and half his. I'm posting it because I need the points a little more than he does - is there a way I can donate half of them to him?)

Define $M$, $I_n$ and $\alpha_n$ as you did. As @Motyla Noga Tomka Mazura points out, the Cauchy-Schwarz inequality shows that $$I_n\le\left(I_{n-1}I_{n+1}\right)^{1/2},$$ hence the sequence $\alpha_n$ is increasing. Hence $\lim\alpha_n$ exists, and hence it's enough to show that $$\lim_{n\to\infty}I_n^{1/n}=M.$$An this is easier than I thought.

First, it's clear that $f^n\le M^n$, hence $I_n\le(b-a)M^n$, so $$I_n^{1/n}\le(b-a)^{1/n}M\to M.$$ So $\limsup I_n^{1/n}\le M$.

Now let $\epsilon>0$. Since $M=\sup f$ and $f$ is continuous, there exists an open interval $E$ such that $$f(t)>(1-\epsilon)M\quad(t\in E).$$If $L$ is the length of $E$ then $$I_n\ge\int_Ef^n\ge L((1-\epsilon)M)^n,$$ so $$I_n^{1/n}\ge L^{1/n}(1-\epsilon)M\to(1-\epsilon)M.$$ Hence $\liminf I_n^{1/n}\ge(1-\epsilon)M$, and hence $\liminf I_n^{1/n}\ge M$.

Use the fact $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n } =g \Longrightarrow \lim_{n\to\infty} \sqrt[n]{a_n } =g$$

WLOG, assume the maximum value of $f$ is $1.$ Then $f^{n+1} \le f^n,$ so the ratio of integrals is $\le 1$ for all $n.$ On the other hand, Holder shows

$$\int_a^b f^n \le (\int_a^b f^{n+1})^{n/(n+1)}\cdot(b-a)^{1/(n+1)}$$

by Holder. Use this and simplify to see

$$(1)\,\,\,\,\frac{(\int_a^b f^{n+1})^{1/(n+1)}}{(b-a)^{1/(n+1)}}\le \frac{\int_a^b f^{n+1}}{\int_a^b f^n} \le 1.$$

But as is well known, $\lim_{p\to \infty}(\int_a^b f^p)^{1/p} = \sup_{[a,b]}|f| = 1.$ It follows that the left side of $(1) \to 1,$ and we're done.