$f$ entire, $g(x, y) := |f(x+iy)|$ integrable on $\mathbb{R}^{2}$ $\implies$ $f \equiv 0$. [duplicate]

Let $$f \in H(\mathbb{C})$$ and $$g(x, y) := |f(x+iy)|$$, and assume $$g$$ is integrable on $$\mathbb{R}^{2}$$. Show then that $$f \equiv 0$$.

Things I Know: $$f$$ is bounded on each $$D(0, R)$$ for $$R > 0$$, but the bounds must increase as $$R \rightarrow \infty$$. Intuitively, this should mean $$\int_{\partial D(0, R)} |f(z)| dz \rightarrow \infty$$ as $$R \rightarrow \infty$$ (assuming $$f \not\equiv 0$$), which could say something about $$g$$. But here I'm talking about upper bounds, which don't help. I could consider the infimum $$f$$ takes on $$\partial D(0, r)$$ to bound below, but these inf's don't necessarily grow larger as $$R$$ does.

My intuition tells me Liouville will come into play, but I'm not sure... Any ideas? Thank you!

Edit: Though this is the same as this question essentially, I want to include proofs that use any complex analytic tools, while the linked question wants a solution based only in elementary calculus.

By the mean-value property of harmonic functions, $$f(z)$$ is the average of $$f$$ over any circle centred at $$z$$, and therefore over any disk centred at $$z$$. A disk $$D$$ of radius $$r$$ centred at $$z$$ has area $$\pi r^2$$, so $$\int_D |f(x+iy)|\; dx\; dy \ge |f(z)| \pi r^2$$ and as $$r \to \infty$$ the only way to avoid an infinite limit is $$f(z)=0$$.