The Hardy-Littlewood maximal operator $$Mf(x)=\sup_{x\in B}\frac1{\vert B\vert}\int_B\vert f(y)\vert dy$$ where the supremum is taken over all balls $B\subset\mathbb{R}^n$ which contain $x$.

It is well-known that $M$ is both strong (for $p>1$) and weak-type (for $p\geq1$) integral operator; see for example this paper and references therein. It's a very nice operator.

Now, replace $M$ by a weighted-Maximal operator $$M_{\mu}f(x)=\sup_{x\in B}\frac1{\mu(B)}\int_B\vert f(y)\vert \,d\mu(y)$$ where the Gaussian measure $d\mu(y)=e^{-\vert y\vert^2/2}dy$ takes over the Lebesgue measure.

Things become bad. I believe the following should be true. Any idea for a proof or reference?

The operator $M_{\mu}$ in not of weak-type $(p,p)$ for any $p\geq1$.