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Assume that $h_\lambda(x)=\frac{1}{\pi} \frac{\lambda}{\lambda^2+x^2}$, for $\lambda>0$, $x \in \mathbb{R}$.

I know that if $f\in L^p$ then $\lim_{\lambda \rightarrow 0} \|f*h_\lambda -f\|_p =0$, for $1\leq p< \infty$ ( Rudin, Real and complex analysis, Thr.9.10).

How to prove that if $f\in L^1$ then $\lim_{\lambda \rightarrow 0} f*h_\lambda (x)=f(x) \textrm{ a.e. ?}$

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    @AlexBecker no it does not. You can get convergence in measure along any particular sequence from that condition, but not a.e. unless you know something else about the form of the kernel. In this case, since it satisfies a weak-type maximal inequality you can show that it differentiates $L^p$.2012-03-18

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