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I know that if $T(X) = f(W(X))$ for one-to-one $f$, where $W(X)$ is minimal sufficient, then $T(X)$ is also minimal sufficient. But my textbook does not include "almost everywhere" or "almost surely" in this statement, which I think must be a typo?

For example, suppose we have $W(X)$ as being minimal sufficient for $\theta$, and we also have that for some measurable function $f$,

$$P_\theta (T(X) = f(W(X))) = 1$$

Is that enough to conclude $T(X)$ is minimal sufficient? After all, if it only fails in null sets, it doesn't seem to be of importance.

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  • $\begingroup$ This is close to nit-picking: the random variable $X$ is defined (as a function) almost everywhere rather than everywhere so the answer is yes. $\endgroup$ – Xi'an Oct 28 '18 at 16:19

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