Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

2 Answers 2

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 $$ \mu\left(X\neq X'\right)=\mu\left(\bigcup_{k\in\mathbb{N}} \left\{|X-X'|>\tfrac{1}{k}\right\} \right)\leq\sum_{k\in\mathbb{N}}\mu\left(|X-X'|>\tfrac{1}{k}\right)=0. $$

share|improve this answer
    
@ Stefan, Could you please explain how the last line was deduced? What is k's role here? –  tear_drops Oct 19 '12 at 3:05
    
@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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.