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I've a question about the definition of a property almost everywhere. On the German Wikipedia site, the definition is:

Let $(X,\mathcal{A},\mu)$ be a measured space. A property $A$ holds almost everywhere in $X$ if and only if $\exists N\in \mathcal{A}$ with $\mu(N)=0$ and $A$ is true for all $x\in X\backslash N$.

The next sentences confuses me (I translate):"Note, the set where $A$ does not hold, need not to be measurable."

But I thought, we can deduce from the definition: The set where $A$ does not hold has measure $0$.

For example suppose we know that a measurable function is almost everywhere discontinuous. Now let $B:=\{x\in X; f \mbox{ is continuous in }x\}$. Is the measure of $B$ equal zero?

Thanks for your help


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For example, take $X$ which is the real line with Borel sigma-algebra and Lebesgue measure, $C$ the Cantor ternary set and $N$ a non-measurable subset of $C$. The property $A$ defined by $x\notin N$ is true for all $x\in X\setminus N$, but the set where $A$ doesn't hold is $N$ which is not measurable. But it's contained in a measurable set of measure $0$ (your can take $N':=C$, the property is still true for all $x\in X\setminus C$). In your example, the set which contains the points of continuity of a function is a countable intersection of open sets, hence measurable. – Davide Giraudo Mar 7 '12 at 10:19

2 Answers 2

up vote 6 down vote accepted

This is true if the measure space is complete, that is if every subset of a measurable set with measure zero is measurable too.

In general, it can be that you have a measurable set $N$ with $\mu(N)=0$ and a nonmeasurable set $M\subseteq N$. If a property holds exactly for the elements of $X\backslash M$, it also holds for every element in $X\backslash N$ and therefore almost everywhere, but the set of elements on which it holds, $X\backslash M$ is not measurable.

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Adding to Michael's answer, a sufficient condition for the measure space $(X, \mathcal{A}, \mu ) $ to be complete is if it is the result of a Carathéodory construction from an outer measure $\mu^*$ on $X$ (this is the case with the Lebesgue measure on $X = \mathbb{R}^n$).

Proof: From the Carathéodory construction we have $$ \mathcal{A} = \{ U \subseteq X : \mu^*(S \cap U) + \mu^* (S\cap U^c) = \mu^*(S), \text{ for all } S \subseteq X \} .$$

Let $A$ be a measurable set with zero measure, so that $\mu^*(A) = \mu(A)=0.$ By the monotonicity property of out measures we have $ \mu^*(N) \leq \mu^* (A)$ for all $N \subseteq A,$ so $\mu^*(N)=0.$

For any $S \subseteq X$, we have $\mu^*(S \cap N)\leq \mu^*(N) =0$ and $ \mu^*(S \cap N^c) \leq \mu^*(S) $ again by the monotonicity property, so $$\mu^*(S \cap N) + \mu^* (S\cap N^c) \leq \mu^*(S).$$

On the other hand, $(S \cap N)\cup (S \cap N^c)= S $ so $$ \mu^*(S) \leq \mu^*(S \cap N) + \mu^* (S\cap N^c) .$$

Thus $\mu^*(S) = \mu^*(S \cap N) + \mu^* (S\cap N^c) $, so $N \in \mathcal{A}.$ That is, $N$ is measurable.

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