$L^1$-integrability

Question

My question comes from some problems I had with the boundedness of the Hilbert transform. In any case:

If a function $f$ is weak-$L^1(\mathbb{R})$ and $L^2(\mathbb{R})$, is it possible to say that it is also $L^1(\mathbb{R})$?

Thanks a lot.

Answer 1

Consider $f(x)=\frac{1}{x}\chi_{[1,\infty)}(x)$, which is $L^2$ but not $L^1$ because of the singularity at $\infty$. But $$|\{x\mid |f(x)|>\frac{1}{\lambda}\}\leq\frac{1}{\lambda}$$ so $f$ is w-$L^1$.