Separability of a space implies existence of an element with positive measure neighborhood.

Question

Suppose $E$ is a separable Hilbert space and $X$ and $X'$ are measurable functions from $(\Omega,\mathcal{F},\mathbb{P})$ into $E$ such that $$\mathbb{P}\left( X-X'\neq 0 \right)>0.$$

I want to show that there exists $a\in E$ such that $$\mathbb{P}\left(\|X-X'-a\|<\frac{1}{3}\|a\| \right)>0.$$

I honestly have no idea... Any help would be appreciated. Thanks in advance.

Answer 1

Let $(a_i)$ be a countable dense set. If $y \in E$ and $\|y-a_i\| \geq \frac 1 3 \|a_i\|$ for all $i$ then $y=0$: Just take a subsequence of $(a_i)$ converging to $y$.

Now prove the result by contradiction. If the claim is false then $P(\|X-X'-a\| < \frac 1 3 \|a_i\|) =0$ for all $i$. Hence there is one null set $S$ outside which we have $\|X-X'-a_i\| \geq \frac 1 3 \|a_i\|$ for all $i$ which implies that $X-X'=0$ with probability $1$.