# What does a.s.P. mean in this context?

I am reading a statistics book, "Sufficient Dimension Reduction" by Bing Li. In Chapter 2, it says:

We say that $$\mathcal{G}_1$$ and $$\mathcal{G}_2$$ are conditional independent given $$\mathcal{G}_3$$, ...,if for every $$A \in \mathcal{G}_1$$ and $$B \in \mathcal{G}_2$$, we have $$P(A \cap B | \mathcal{G}_3)=P(A|\mathcal{G}_3) P(B | \mathcal{G}_3), a.s.P.$$

What does a.s.P. mean in this context? I didn't find clues in previous texts, so I guess this should be a pretty common symbol.

• math.stackexchange.com/questions/731283 Commented May 3, 2022 at 16:20
• A property $\mathcal{P}$ holds $P$-a.s. if the set $B:=\{x:\mathcal{P}(x)=false\}$ has $P$ measure $0$, that is $P[A]=0$. Commented May 3, 2022 at 16:24

Two measurable functions $$f,g$$ are said to be equal $$\mu$$-almost everywhere (shorthand: $$\mu$$-a.e.) if $$\mu (\{x \, : \, f(x) \neq g(x)\}) = 0$$. In probability theory, the more common terminology is "almost surely" (a.s.) instead of almost everywhere. If one wants to specify the probability measure, one can say "almost surely for $$P$$" (a.s.$$P$$.).
In your case, notice that $$P(A \cap B | \mathcal{G}_3)$$, $$P(A|\mathcal{G}_3)$$, and $$P(B|\mathcal{G}_3)$$ are each random variables (i.e. measurable functions on a probability space), hence the statement means that for the set:
$$T = \{ \omega \in \Omega \, : \, P(A \cap B | \mathcal{G}_3)(\omega) \neq P(A|\mathcal{G}_3)(\omega) P(B|\mathcal{G}_3)(\omega)\}$$ we have $$P(T) = 0$$.