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I'm trying to follow a proof for showing $\displaystyle \lim_{n\rightarrow \infty} P[|X_n-X|>\epsilon] = 0 \Rightarrow X_n \rightarrow_p X$

The first step of the proof says:

$P[X \leq x-\epsilon] - P[|X_n - X| \geq \epsilon] \leq P[X_n \leq x] \leq P[X\leq x+\epsilon] + P[|X_n - x| \geq \epsilon]$

The first inequality is true because any $\omega$ such that $X(\omega) \leq x-\epsilon$ and $X_n - X \leq \epsilon$, will certainly satisfy $X_n(\omega) \leq x$.

But I cannot think of similar reasoning to see why the latter inequality is true.

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I believe there is a significant typo in the final term making the second inequality impossible to prove.

Claim: $P[|X_n - x| \geq \epsilon]$ should be $P[|X_n - X| \geq \epsilon]$.

Proof of claim:

Let $\Omega$ have only one element: $\omega$. In that case, $\mu(\omega) =1$, and $X_n = X_n(\omega)$ is just a sequence of real numbers. Suppose it is defined by: $$ X_n = \begin{cases} n & n \leq 10 \\ 10 & n > 10 \end{cases} $$ Also, let $x = 5$, $n = 10$, and $\epsilon = 1$. Then $$P[X_5 \leq 5] = 1$$ because in fact $P[X_5 = 5] = 1$. Therefore, $$P[|X_5 - 5| \geq 1] = 0.$$ But we know that since the sequence $X_n \to 10$, we have the fact that $X = 10$. And hence, $$P[X \leq 5+1] =0.$$ This contradicts your second inequality because $$ 1> 0 +0.$$

Proof with typo fixed:

Let $A = \{\omega : X_n(\omega) \leq x\}$, $B = \{\omega :X(\omega) \leq x+\epsilon\}$, and $C = \{\omega : |X_n(\omega) - X| \geq \epsilon\}$.

What we want to show is $P(A) \leq P(B) + P(C)$. It suffices to show $A \subseteq B \cup C$. So let $\omega \in A$, and assume that $\omega \notin C$. We have these two inequalities:

  • $X_n(\omega) \leq x$
  • $X(\omega) - X_n(\omega) < \epsilon$

By adding them we see that they imply that $X(\omega) \leq x + \epsilon$. Therefore $\omega \in B$. We have just shown that $A \subseteq B \cup C$, completing our proof of the second inequality.

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  • $\begingroup$ Thanks -- there was a typo in the book. $\endgroup$ – MathStudent Jul 8 '14 at 19:51
  • $\begingroup$ You're welcome. (The ironic thing is that as I copied down what you wrote, I accidentally wrote a capital X! After that, I rechecked your post and then noticed the typo.) $\endgroup$ – Andrew Kelley Jul 8 '14 at 20:09

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