# Lebesgue integral over "bad" measurable sets

Let $\Omega \subset \mathbb{R}^n$ ($n \geq 1$) be a bounded open domain and $f \in L^\infty(\Omega)$ possibly changes the sign. Assume that the set $$\Omega^+ := \{x \in \Omega: f(x) > 0 \}$$ has positive measure. Is it true that there exists non-negative $\varphi \in C_0^1(\Omega)$, such that $$\int_\Omega f \, \varphi \, dx > 0 ~?$$ Obviously, if $\Omega^+$ has nonempty interior (after redefinition on a set of measure $0$), then the answer is Yes. But I'm afraid of the case $\Omega^+$ is a "bad" set, in the sense that its interior is empty (like fat Cantor set).

Thanks.

• $C^1_0$ means differentiable with compact support?
– fgp
Mar 30 '14 at 12:47
• @fgp Yes, just differentiable with compact support. Mar 30 '14 at 12:51

For $g(x) = \chi_{\Omega^+}$, it is clear that $$\int_{\Omega} f g = \int_{\Omega^+}f = R> 0 \text{,}$$ but of course in general $g \notin C^1_0$. However, since $\Omega$ is bounded, $g \in L^1(\Omega)$ because $\int_{\Omega} g \leq \int_{\Omega}1 < \infty$. Since $C^1_0(\Omega)$ is a dense subset of $L^1(\Omega)$, there is thus some $\varphi \in C^1_0(\Omega)$ with $$\|\varphi - g\|_1 < \frac{R}{2\|f\|_\infty}$$ and thus with $$\int f(\varphi - g) \leq \|f\|_\infty\cdot\|\varphi - g\|_1 \leq \frac{R}{2} \text{.}$$ But then $$\int_{\Omega} f\varphi = \int_{\Omega} fg + \int_{\Omega} f(\varphi - g) \geq R - \frac{R}{2} = \frac{R}{2} > 0 \text{.}$$