Meaning of "almost everywhere" in measure theory.

I'm slightly confused about the term almost everywhere as it is used in Folland's real analysis.

Given a measure space $(X, \mathcal{M}, \mu)$ Suppose $f \equiv g$, $\mu$-almost everywhere where $f, g : X \to \mathbb{R}$.

Does this mean that $$\mu(\{x : f(x) \ne g(x) \}) = 0$$

Or that there exists a measurable set $E$ such that $\{x : f(x) \ne g(x) \} \subseteq E$ and $\mu(E) = 0$?

This issue came up when my professor was proving the following theorem from Folland:

To prove (a) $\implies$ $\mu$ is complete, he took a null set $N \in \mathcal{M}$ and said for any $E \subseteq N$, $1_E \equiv 0$ almost everywhere. This part confused me, because how can we know if $E$ is measurable?

• In most texts I've seen, 'almost everywhere' means all but in a set of measure zero. I'd expect the set you mention is in fact measurable since all non-measurable sets only come up with the axiom of choice used, but probably both suffice for most uses. Mar 3, 2014 at 3:54
• @Hayden Your last comment is only true if the measure being discussed is Lebesgue measure, which is not the case here. Mar 3, 2014 at 4:10
• The meaning of a.e. is given in page 26: A statement is true $\mu$-a.e. iff it holds except for those $x$ in some null set. Although this is not quite explicit there, the idea is that if $N$ is null and a property holds in $X\setminus N$, then it holds a.e., even if it also holds for some $x\in N$, and the actual set of exceptions is not measurable. Mar 3, 2014 at 4:20
• That makes sense. I guess when reading things like $P(x)$ holds for all $x$ except for $x \in V$, I interpret that as $V$ is the set of all points for which $P(x)$ doesn't hold. Mathematical language can be confusing. Mar 3, 2014 at 4:25
• Rudin defines almost everywhere as meaning the set where something doesn't happen is a subset of a set of measure 0. I already answered the same question but can't find it. May 8, 2017 at 15:02

Although the comments more or less answer the question, and the question is a few years old, some readers might benefit from a complete answer.

Here is another way to think about the concept. "Almost everywhere" means some proposition/property is true everywhere except a set of measure zero. So if $f=g$ a.e. this means the following.

Let $f,g:X\rightarrow Y$ and let $E\subset X$ with $\mu(E)=0$. If $f=g$ on $X\setminus E$ and $f\neq g$ on $E$ then $f=g$ a.e.

Note that here, we did not assume the Lebesgue measure, or that we are even dealing with domains which are subsets of $\mathbb{R}^n$.

Here are some good examples of what can "go wrong" with functions equal almost everywhere. For simplicity, let's consider the Lebesgue measure on $\mathbb{R}$

$(1)$ Let $D(x)$ denote the Dirichlet function, where $D(x)=0$ if $x$ irrational and $D(x)=1$ if $x$ is rational. Now consider the zero function $f(x)=0$. Note that $f=D$ almost everywhere, since $\mu(\mathbb{Q})=0$ and $f\neq D$ only on $\mathbb{Q}$.

So even if two functions are equal almost everywhere, one can be continuous everywhere and the second one can be nowhere continuous.

$(2)$ Cantor's function is constant a.e. but monotone increasing.

$(3)$ Define $g(x)=0$ if $x$ irrational and $g(x)=x$ if $x$ is rational.

$g(x)$ is bounded a.e. but not globally bounded.

In addition, you can show two functions in $L^2$ are equal almost everywhere if you can show their Fourier series and/or Fourier transforms are equal almost everywhere. There are countless of other such examples.

• It always struck me as odd that functions can be equal almost everywhere, yet disagree on a dense subset (viz. (1)). May 8, 2017 at 15:44
• Agreed. I can only imagine what state the mathematical community was in when Lebesgue introduced the concept of "measure", and they began toying around with the concept of almost everywhere. May 8, 2017 at 16:51