# Problem with Sequences in Measures

If ${X_n \to X}$ and ${X_n \to X'}$ in measure, then show that $\mu(X \neq X') =0$.

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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 $$\mu\left(X\neq X'\right)=\mu\left(\bigcup_{k\in\mathbb{N}} \left\{|X-X'|>\tfrac{1}{k}\right\} \right)\leq\sum_{k\in\mathbb{N}}\mu\left(|X-X'|>\tfrac{1}{k}\right)=0.$$
@tear_drops: We want to express the set $\{X\neq X'\}$ as a countable union in order to get the last inequality (this inequality will not hold in general if it were an uncountable union). Then it is a standard trick to show that $\{X\neq X'\}$ is equal to $\bigcup_k \{|X-X'|>\tfrac{1}{k}\}$. – Stefan Hansen Oct 19 '12 at 5:21
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.