$f$ is an entire function such that$$I = \int_{\mathbb{R}^2} \mid f(x + iy \mid dx dy < \infty,$$ then show that $f = 0$.
Case I: If $f$ is a non-zero constant function, then $I = \infty$. Hence false case.
Case II: If $f$ in non-constant, then $f$ is unbounded. But there is still a possibility that $\mid f \mid \geq M$ on a set of zero measure. In this regard I think I can make use of a theorem which says that Im$(f)$ is dense in $\mathbb{C}$. So I think can overcome the earlier objection and show $\mid f \mid \geq M$ has a non-zero measue. Hence if this is true then again this case is false.
Therefore the only case left is $f = 0$.
So can someone verify my logic. Also I not able to mathematically write my intuition in case II, so I need help there.
This is a corollary of Liouville's theore. Let $c = \int_{\mathbb{C}}|f(w)|\,dV(w)$, where $dV(w) = \,dx\,dy$. We will show that your condition implies that $f$ is bounded. Pick a point $z \in \mathbb{C}$ and consider the ball centered at $z$ with radius $1$. By the mean value property for holomorphic functions we have \begin{equation*} f(z) = \frac{1}{\pi}\int_{B(0,1)}f(w)\,dV(w) \end{equation*} Thus, \begin{equation*} |f(z)| \leq \frac{1}{\pi}\int_{B(0,1)}|f(w)|\,dV(w) \leq \frac{c}{\pi} \end{equation*} This implies $f$ is bounded and hence a constant. The only constant integrable function is $f \equiv 0$.