Prove for two random variables if $P(|X-Y|\geq\epsilon) = 0$, then they are equal Problem statement: X and Y are random variables and $\epsilon > 0$. If
$P(\{|X-Y|\geq\epsilon\}) = 0$
prove that
$P(\{X=Y\}) = 1$.
The straightforward solution is breaking the probability of the inequality into sum of disjoint probabilities. My question is whether is attempt below using the convergence of probability is correct.
Attempt: Consider sequence of random variables $X_n$ such that $X_n = X$ for all $n$. Then,
$P(\{|X_n-Y|\geq\epsilon\}) = 0 \\
\Rightarrow \lim\limits_{n \to \infty} P(\{|X_n-Y|\geq\epsilon\}) = 0 \\ 
\Rightarrow X_n \overset{Pr}\rightarrow Y  \\
\Rightarrow X \overset{Pr}\rightarrow Y$
So $X$ converges in probability to $Y$. Since $X$ is not a sequence I want to conclude something like $X=Y$ but not sure if this approach is correct.
 A: We have $\forall \epsilon >0, \hspace{0.1cm} \mathbb{P}(|X-Y| < \epsilon) = 1$. Then by letting $\epsilon \to 0$ and using the continuity the probability we have $\mathbb{P}(|X-Y| \le 0) = 1$ so $\mathbb{P}(X = Y) = 1$.
Concerning your approach you should show that $X(w) = Y(w)$ for all $w \in \Omega$. Obviously this not a direct implication of the convergence in probability
A: This follows easily from Markov's inequality. Let the indicator $Z = I_{|X - Y| \ge \epsilon}$.
$$ P(Z \ge \epsilon) \le \frac{P(|X-Y| \ge \epsilon)}{\epsilon} = 0 $$
implies that $P(Z \ge \epsilon) = 0$ for all $\epsilon >0$. Hence $P(Z = 0) = 1$, which implies that $|X-Y| = 0$.
A: First, I think the statement should be 'If for all $\epsilon>0$, $P(|X-Y|\geq \epsilon)=0$ then $P(X=Y)=1$. For example, $X$ and $Y=X+1$ satisfy this for $\epsilon=2$ but obviously is not true.
Now look at your moving from your 2nd line to your third? Why is this true? With $X_n=X+1$ and $\epsilon=2$ this is true but $X+1$ does not converge to $X$.
I would consider looking at $\Omega_n=\{\omega: |X(\omega)-Y(\omega)|\leq \frac{1}{n}\}$. $\Omega_{n+1}\subset \Omega_n$, $P(\Omega_n)=1$, and $\cap_n\Omega_n=\{\omega:X=Y\}$. What is $P(\cap_n\Omega_n)$? Have you seen a theorem whose conditions are similar to this?
