Problem with Sequences in Measures If ${X_n \to X} $ and ${X_n \to X'}$ in measure, then show that $ \mu(X \neq X') =0 $.
 A: Fix $\varepsilon>0$ and $k\geq 0$; then for $n\geq N(k,\varepsilon)$, $\mu(|X_n-X|\geq\varepsilon)\leq 2^{-k}$ and $\mu(|X_n-X'|\geq\varepsilon)\leq 2^{-k}$ so for such $n$, by triangular inequality,$$\mu(|X-X'|\geq 2\varepsilon)\leq\mu(|X_n-X|\geq \varepsilon)+\mu(|X_n-X'|\geq \varepsilon)\leq 2\cdot 2^{-k}.$$ As $k$ is arbitrary we get that $\mu(|X'-X|\geq j^{-1})=0$ for each $j>0$, hence $X=X'$ almost surely.
An other way to see that, but which uses a deeper argument, it to notice and show that $X_n$ converges in measure to $X$ if and only if $E\frac{|X_n-X|}{1+|X_n-X|}\to 0$ (see here for example). This corresponds to a metric, hence the limit is necessarily unique.  
A: Let $\varepsilon >0$ be given. Then the triangle inequality yields that
$$
\{|X-X'|>\varepsilon\}\subseteq \{|X-X_n|>\tfrac{\varepsilon}{2}\}\cup \{|X'-X_n|>\tfrac{\varepsilon}{2}\}
$$
holds for every $n\in\mathbb{N}$ (to convince yourself, you can look at the complements). This implies that
$$
\mu\left(|X-X'|>\varepsilon\right)\leq \mu\left(|X-X_n|>\tfrac{\varepsilon}{2}\right)+\mu\left(|X'-X_n|>\tfrac{\varepsilon}{2}\right)
$$
holds for every $n\in\mathbb{N}$ and by the assumption we have
$$
\mu\left(|X-X'|>\varepsilon\right)\leq \lim_{n\to\infty}\left[\mu\left(|X-X_n|>\tfrac{\varepsilon}{2}\right)+\mu\left(|X'-X_n|>\tfrac{\varepsilon}{2}\right)\right]=0.
$$
Thus
$$
\mu\left(X\neq X'\right)=\mu\left(\bigcup_{k\in\mathbb{N}} \left\{|X-X'|>\tfrac{1}{k}\right\} \right)\leq\sum_{k\in\mathbb{N}}\mu\left(|X-X'|>\tfrac{1}{k}\right)=0.
$$
