Let $\mathcal{R}$ be the set of rectangles in the plane and, given $f \in L^1$ let $$ f^*(x) = \sup_{x \in R \in \mathcal{R}} \frac{1}{ \lvert R \rvert} \int_R \lvert \, f \,\rvert $$ as defined in this question. You can show that the weak-type inequality fails for this operator, that is, there is no constant $A$ such that $$ m(\{ f^* > \alpha \}) < \frac{A}{\alpha} \lVert \, f \, \rVert_1 $$ for each integrable $f$. From Stein and Shakarchi's book on real analysis I'd like to show the following, much stronger claim: there is an integrable $f$ such that
$$ \limsup_{\text{diam}(R) \to 0} \frac{1}{|\,R\,|} \int_R \lvert \, f \, \rvert = \infty \text{ a.e.} $$ where diam, of course, is the diameter. According to the book this "should" follow from the failure of the weak type inequality, but I don't really know how to do it.
I was thinking something like the following, but it failed. It suffices to find and $f$ such that the desired conclusion holds only on a set of positive measure, for we can translate around and get the desired effect. So I wanted to show that if the $\limsup$ expression is finite everywhere, the weak-type inequality holds, for that specific function. That would get me what I want. However I got nowhere with finding a function for which the weak-type inequality fails, sadly.
Edit: Whoops, this is actually very hard. In its most general form it relies on a result found on page 441 of Stein's book Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory Integrals. Hopefully this specific case can be had more easily, though?