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Let $f:\mathbb{R}\to\mathbb{R}$ be locally integrable and define a function $\rho_L$ for $L>0$ by $\rho_L(x) := \frac1{x\ln L} 1_{[1/L,1]}(x).$ Obviously $\int\rho_L = 1$. Is it true that the convolution $f*\rho_L(x)$ converges to $f(x)$ in the $L\to\infty$ limit almost everywhere? The fact that this $\rho_L$ is not a approximation to the identity (i.e., not a scaled version of some $\rho_1$) makes the standard proofs (using Hardy-Littlewood's maximal function) not applicable, it seems to me.

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    For continuous functions it is true. Just split the integral up into two pieces at the point $L^{-\alpha}$ for \alpha>0. The piece close to the singularity will give you $(1-\alpha)f(x)$ as $L\to \infty$ and the other piece is bounded above by $\alpha \|f\|_\infty$. Then send $\alpha\to 0$. Not sure about locally integrable, but you can probably use a density argument or something.2012-09-20

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