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I've recently started learning Lebesgue integral and having come across this problem I'm not sure how one would prove this.

Let $f(x):[a,b] \to \mathbb{R}$ be continuous and let $M=sup_{x\in[a,b]}|f(x)|$. suppose that $M>0$ and $p>0$.

  • prove that for every $\epsilon$ such that $0<\epsilon<\frac{M}{2}$ there is a non-empty open interval $I \subset [a,b]$ such that $$(M-\epsilon)^p |I| \leq \int_{a}^{b} |f(x)|^p \, dx \leq M^p (b-a)$$
  • deduce that $\lim_{p \to \infty} (\int_{a}^{b} |f(x)|^p \, dx =M$

any help is appreciated.

Shima Shargy
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    Intuitively, there is some point $x_0\in (a,b)$ where $|f(x_0)|>M-\epsilon$. Since $f$ is continuous, it can't be just one point, but there is some interval around $x_0$ such that $|f(x)|>M-\epsilon$ in that interval. Now use the monotonicity of integral. (try to turn this idea into a formal proof) – Mark Nov 08 '21 at 19:31

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