Does $\int_a^bf^p=0$ imply $\int_a^bf=0$?

Question

Let $f:[a,b]\to\mathbb{R}$ be a nonnegative Riemann integrable function. If $\int_a^bf^p=0$ for some $p>0$, is it true that $\int_a^bf=0$ ? It is obviously true if $p=1$, and the Hölder's inequality $$\left|\int_a^bfg\right|\leq\left(\int_a^b|f|^p\right)^{1/p}\left(\int_a^b|g|^p\right)^{1/p}$$ implies that it is true if $p>1$. But what if $0<p<1$ ? I would prefer a solution which does not use Lebesgue integral or measure zero sets if possible, please.

Answer 1

Suppose $ 0<p<1$. Let $M=\sup \{f(x): a \leq x \leq b\}$. Let $g=\frac f M$. Then $\int g^{p}=0$ and $0\leq g \leq 1$. Hence $0 \leq g \leq g^{p}$ which gives $0\leq \int g \leq \int g^{p}=0$ which gives $\int g=0$. This implies $\int f=0$.

Answer 2

First of all, let sign $\ =\ $ mean "almost equal" above.

The Answer is YES.

Otherwise, if integral $\ \int\ $ on the right were positive, there would exist $\ s>0\ $ such that the measure

$$ m\ :=\ \mu(f^{-1}((s;\infty))\ >\ 0 $$

would be indeed positive. Then

$$ \int_a^b f^p\ \ge\ s^p\cdot m\ >\ 0 $$

-- a contradiction which proves "YES".

Answer 3

Here, we show the contrapositive: if $\int_E|f(x)|\,\mathrm{d}x\gt0$, then $\int_E|f(x)|^p\,\mathrm{d}x\gt0$.

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Using the formula $$ \int_E|f(x)|^p\,\mathrm{d}x=\int_0^\infty\mu\{x\in E:|f(x)|\gt\lambda\}\,p\lambda^{p-1}\,\mathrm{d}\lambda\tag1 $$ if $\int_E|f(x)|\,\mathrm{d}x\gt0$, then $$ \int_0^\infty\mu\{x\in E:|f(x)|\gt\lambda\}\,\mathrm{d}\lambda\gt0\tag2 $$ and so there must be some $\lambda_0\gt0$ so that $\mu\{x\in E:|f(x)|\gt\lambda_0\}\gt0$. Consequently, since $\mu\{x\in E:|f(x)|\gt\lambda\}$ is a non-increasing function of $\lambda$, $$ \begin{align} \int_E|f(x)|^p\,\mathrm{d}x &\ge\int_0^{\lambda_0}\mu\{x\in E:|f(x)|\gt\lambda_0\}\,p\lambda^{p-1}\,\mathrm{d}\lambda\\[6pt] &=\mu\{x\in E:|f(x)|\gt\lambda_0\}\,\lambda_0^p\\[9pt] &\gt0\tag3 \end{align} $$