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Let $X, Y, Z:\Omega\longrightarrow \mathbb{R}$ be random variables on a probability space $(\Omega,\mathscr{F},\mathbb{P})$ such that $X$ is independent of $Y$ and $Y = Z$ almost surely/everywhere.

Is it true that, $X$ would also be independent of $Z$?...And if not, are there any further conditions/assumptions that this may then hold true?

I have been trying to prove this for a while because I couldn't come up with a simple counter example (I must admit I am not too great at creating them).

If true, a proof preferably using the $\sigma$-algebra definition of independence between random variables would be amazing as this is something I feel is easier for me to understand, although any other arguement would still be great. Many thanks in advanced.

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    $\begingroup$ Yes, no further conditions. For measurable $A\subset \mathbb R^2$, the event $[ (X,Y)\in A]$ differs from the event $[(X,Y)\in A]$ by a null set. $\endgroup$ – kimchi lover Jul 16 '17 at 23:54
  • $\begingroup$ Sorry what is the difference between the two events that you have written? I do understand that $X$ and $Y$ only differ by a null set....but I still don't see how this implies that $X$ is independent of $Z$. $\endgroup$ – User086688 Jul 16 '17 at 23:56
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    $\begingroup$ $\Pr(X\le x, Z\le y) = \Pr(X\le x, Y\le y) = \Pr(X\le x) \Pr(Y\le y) = \Pr(X\le x) \Pr(Z\le y)$ $\endgroup$ – user251257 Jul 16 '17 at 23:56
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    $\begingroup$ @User086688 Whoops. I meant to write, $[(X,Y)\in A]$ differs from $[(X,Z)\in A]$ by a null set. Which is what the other commenter used, with $A=(-\infty,x]\times(-\infty,y]$. $\endgroup$ – kimchi lover Jul 17 '17 at 0:05
  • $\begingroup$ Ah ok, thanks so much! Got it. $\endgroup$ – User086688 Jul 17 '17 at 0:20

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