1
$\begingroup$

On Wikipedia's entry of Convergence of Random Variables, it states that provided a probability space is complete:

If $X_n\ \xrightarrow{p}\ X$ and $X_n\ \xrightarrow{p}\ Y$, then $X=Y$ (almost surely).

However, I thought that this normally holds for the Borel measureable set instead of the Lebesgue (complete) measure space. Is the entry wrong?

$\endgroup$
1
$\begingroup$

By almost surely it is meant almost everywhere $P$. The entry is right.

For a short proof of this fact, recall that every sequence converging in probability has a subsequence converging almost surely to it's limit. Applying this fact twice in a row we get $X_n\rightarrow X$ a.s. and $X_n\rightarrow Y$ a.s. along the same subsequence. In particular $X=Y$ a.s.

$\endgroup$
  • $\begingroup$ What I don't get is why they mention the probability space being complete. Do the results NOT hold if the space is not complete (ie, Borel measure)? $\endgroup$ – user321627 Jan 15 '17 at 2:22
  • $\begingroup$ The assumption of completeness (i.e., every subset of a null-set is measurable) facilitates proofs since we don't have to worry about "almost sure" sets becoming non-measurable and hence not being able to speak about the probability of such event. $\endgroup$ – mbe Jan 15 '17 at 2:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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