Is the squared difference between a $L^{2}$-function and a Non-$L^{2}$-function in $L^{2}$?

Question

Let $(\Omega, \mathcal{A}, P)$ be a probability space. Furthermore, let $Y,V : \Omega \rightarrow \mathbb{R}^n$ be random vectors and let

$$ \int_{\Omega}{\vert \vert Y \vert \vert ^2}dP < \infty$$ and $$ \int_{\Omega}{\vert \vert V \vert \vert ^2}dP = \infty.$$ Is it true that this implies

$$ \int_{\Omega}{\vert \vert V-Y \vert \vert ^2}dP = \infty?$$

It seems intuitive, but I have a hard time showing it rigorously.

Thanks for any help!

Answer 1

Notice that $\|V\| \leq \|V-Y\| + \|Y\|$ (by the triangle inequality) and so $$\int \|V\|^2 \leq \int (\|V-Y\| + \|Y\|)^2 \leq 4 \int \|V-Y\|^2 + \|Y\|^2$$ so if $\int \|V-Y\|^2 < \infty$ then $\int \|V\|^2 < \infty$, a contradiction.