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:

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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?

  • $\begingroup$ 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. $\endgroup$
    – Hayden
    Mar 3, 2014 at 3:54
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    $\begingroup$ @Hayden Your last comment is only true if the measure being discussed is Lebesgue measure, which is not the case here. $\endgroup$ Mar 3, 2014 at 4:10
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    $\begingroup$ 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. $\endgroup$ Mar 3, 2014 at 4:20
  • $\begingroup$ 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. $\endgroup$
    – user132352
    Mar 3, 2014 at 4:25
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    $\begingroup$ 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. $\endgroup$ May 8, 2017 at 15:02

1 Answer 1


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.

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    $\begingroup$ It always struck me as odd that functions can be equal almost everywhere, yet disagree on a dense subset (viz. (1)). $\endgroup$ May 8, 2017 at 15:44
  • $\begingroup$ 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. $\endgroup$ May 8, 2017 at 16:51

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