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This is the definition of 'almost everywhere' in Folland Real Analysis.

If $(X,\mathcal{M},\mu)$ is a measure space, a set $E \in \mathcal{M}$ such that $\mu(E) = 0$ is called a null set. By subadditivity, any countable union of null sets is a null set, a fact which we shall use frequently. If a statement about points $x\in X$ is true except for $x$ in some null set, we say that it is true almost everywhere (abbreviated a.e.), or for almost every $x$. (If more precision is needed, we shall speak of a $\mu$-null set, or $mu$-almost everywhere.)

But, there is something ambiguous in the def. What I understand is that a statement on X is almost everywhere true if the subset of X in which the given statement is not true is a 'null-set' itself. It seems to be true in the context...but the folland book is written somewhat ambiguously.. Am I correct?

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  • $\begingroup$ What makes you say “ambiguous”? Is there a statement which is true almost everywhere in one interpretation of the above, and not true almost everywhere in another interpretation? $\endgroup$ – Matthew Leingang Jan 20 '15 at 17:22
  • $\begingroup$ Could you clarify what you find ambiguous? You've encapsulated the definition just fine: the set on which it doesn't hold has measure 0, hence "null-set." $\endgroup$ – Alex R. Jan 20 '15 at 17:22
  • $\begingroup$ @AlexR.: See my answer, I think this is the ambiguity the OP speaks about. $\endgroup$ – PhoemueX Jan 20 '15 at 17:24
  • $\begingroup$ The phrase 'except for x in some null set' means that the statement is false for all x in the null set? $\endgroup$ – Keith Jan 20 '15 at 17:37
  • $\begingroup$ @Merideth: The way I interpret it (and this is one of the ambiguities) is that this does not mean that the statement has to be false on the null-set. With this interpretation, Folland's definition coincides with the usual one. $\endgroup$ – PhoemueX Jan 20 '15 at 17:58
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You are not completely correct.

Ordinarily, one says (and I interpret Folland as meaning this) that a property $E(x)$ is valid ($\mu$-)almost everywhere, if there is a ($\mu$-)null set $N \in \mathcal{M}$ such that $E(x)$ is true for all $x \in X \setminus N$.

If the set $M := \{x \in X\mid E(x) \text{ is true}\}$ is measurable, then this is the same as saying that $M^c$ is a null-set.

But it can happen that $M$ is not measurable, although $E(x)$ holds almost everywhere.

For a (trivial) example, consider $X = \Bbb{N}$, $\mathcal{M} = \{\emptyset, X\}$ and $\mu \equiv 0$. Then every property holds almost everywhere (we can take $N = X$), but if $E(x)$ means (e.g.) that $x$ is even, then $M$ is not measurable.

This is somewhat related to the concept of a complete measure space (this is treated in Folland), which is a measure space with the property that if $N \in \mathcal{M}$ has measure $0$ and if $M \subset N$, then $M \in \mathcal{M}$.

The measure space given above is not complete.

In complete measure spaces, both formulations are easily seen to be equivalent.

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    $\begingroup$ These are are very valid points. But, I'm not quite seeing the ambiguity. Afterall, the definition still works (even if it can be stupid in some examples). In other words, could you derive some horrible contradiction without assuming a complete metric space? I think the point of Folland's definition is that apriori, null sets are defined to be measurable and have measure 0. He could have just as well defined $E(X)$ to be measurable, such that $E(X)^c$ has measure 0. $\endgroup$ – Alex R. Jan 20 '15 at 17:35
  • $\begingroup$ In your explanation, do you mean that E(x) can hold for some x in N? $\endgroup$ – Keith Jan 20 '15 at 17:35
  • $\begingroup$ So, the set of all x such that E(x) is 'false' can be a proper subset of N? $\endgroup$ – Keith Jan 20 '15 at 17:37
  • $\begingroup$ @Merideth: Yes, exactly. $\endgroup$ – PhoemueX Jan 20 '15 at 17:54
  • $\begingroup$ @AlexR.: I do not say that there is a contradiction. I interpreted the OP in that he was asking clarification of (among others) the point if the definition implies that the set where $E(x)$ holds (or does not hold) is measurable. This is not the case with the definition given by Folland. In proofs, this just has the implication that one should write "Let $N$ be a null-set such that $E(x)$ holds for all $x \in X\setminus N$" instead of "Let $N = \{x \mid E(x) \text{ is false}\}$." I will let the OP decide in how far my answer actually adresses his question :) $\endgroup$ – PhoemueX Jan 20 '15 at 17:56

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