If $\int_{M} f \ d\mu > 0$ then $f(x) > 0$ for some $x \in M$?

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Let $M \subseteq \mathbb{R}^n$ compact and $f : \mathbb{R}^n \to \mathbb{R}$ a continuous function. Given that the Lebesgue integral $$\int_{M} f \ d\mu = \int_{\mathbb{R}^n} f \mathcal{X}_{M} \ d\mu \ > \ 0$$ where $\mathcal{X}_{M} : \mathbb{R}^n \to \{0,1\}$ is the characteristic function. Does it follow that there exists at least one point $x \in M$ (and therefore by continuity, an open ball $B_{r}(x)$) such that $f(x) > 0$?

It seems intuitively obvious, but I can't prove it. Thanks.

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Yes. Otherwise,$$(\forall x\in M):f(x)\leqslant0$$and therefore$$\int_Mf\,\mathrm d\mu\leqslant\int_M0\,\mathrm d\mu=0.$$