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Let $g:\mathbb{R}\longrightarrow\mathbb{R}$ be a locally integrable function and $\phi\in\mathcal{C}^\infty_c(\mathbb{R})$ (i.e. $\phi $is a $\mathcal{C}^\infty$ function with compact support in $\mathbb{R}$)

Is it true that $g\phi$ is integrable, in the sense of Riemann?

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No. Let $g = \chi_{[0,1] \cap \mathbb Q}$, and $\phi \in C^\infty_c(\mathbb R)$ some function which is $1$ on all of $[0,1]$. The $g\phi = g$ and $g$ is not Riemann-integrable.

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    Thanks. Would it become true if $g = f^\prime$ a.e. , where $f$ is a continuous function defined on $\mathbb{R}$ ?2012-11-04