I suppose I'm a bit confused about the definition in the following regard:

A property holds a.e. if it holds everywhere except for a set of measure $0$. Now, if the particular property is only defined for a set of measure $0$, is it a.e. by default?

Say I have two 'continuous' (standard topology) sequences $f,g: \mathbb{N} \to \mathbb{R}.$ Are we then allowed to say that $f = g$ a.e.? Or instead do the functions have to be defined on a set of non-zero measure and a.e. refers to some measure $0$ subset?

I ask because a homework exercise asks me if two real functions are continuous and agree a.e. on a subset of $\mathbb{R}$, are necessarily identically equal. Clearly this is true if the points are not isolated, since if continuous functions disagree at some point, they must disagree on a non-zero measure set since open sets have non-zero measure. Though it need not be if I just use sequences.

So, what are the requirements to use the phrase a.e.?

Wolfram definition: A property of $X$ is said to hold almost everywhere if the set of points in $X$ where this property fails is contained in a set that has measure zero.

This would seem to imply that it is a.e. by default.

  • $\begingroup$ For your first question I think the answer is quite clear: NO! A point ${x}$ is of mesure zero and if $f(x)=5$ you don't have f constant equal to $5$ $\endgroup$ – Piquito Oct 2 '15 at 20:26
  • $\begingroup$ @Ataulfo I don't see how this answers the question... $\endgroup$ – Anthony Peter Oct 2 '15 at 20:30
  • $\begingroup$ For it to be meaningful for functions $f, g: \mathbb N \rightarrow \mathbb R$ you have to have a measure that is not what you usually use at $\mathbb R$. Or else the measure of $\mathbb N$ will be zero (and almost everywhere could have arbitrary exceptions). And regarding the homework, there may be an ambiguity here. Do they mean they are continuous a.e and equal a.e do they mean they are equal a.e and continuous everywhere? $\endgroup$ – skyking Oct 2 '15 at 20:30
  • $\begingroup$ @skyking Why? The question asks for any subset, this doesn't seem to exclude a subset with measure 0 $\endgroup$ – Anthony Peter Oct 2 '15 at 20:32
  • $\begingroup$ @AnthonyPeter The point is important, because if you allow the functions to be discontinuous on a set of measure zero they can differ there also and still be equal a.e. $\endgroup$ – skyking Oct 2 '15 at 20:35

To speak of a property holding a.e. on a set $X$ you first need a measure. If $P(x)$ is a property defined for every point $x \in X$, then $P$ is true $\mu$-almost everywhere if $$\mu(\{ x \in X \mid \neg P(x)\}) = 0.$$ Perhaps a bit more precisely, you might require that there is a $\mu$-measurable set $N$ with the property that $\mu(N) = 0$ and $\{ x \in X \mid \neg P(x)\} \subset N$ since the first set is possibly not $\mu$-measurable.

What does it mean for two functions $f,g : \mathbb N \to \mathbb R$ to be equal almost everywhere? Without a measure this question can't be answered. If you specify a measure $\mu$ on $\mathbb N$, they are equal $\mu$-a.e. if $$\mu(\{n \in \mathbb N \mid f(n) \not= g(n)\}) = 0.$$ For instance, if $\mu$ is the counting measure, this states the set of points where $f(n) \not= g(n)$ has zero elements - in this case $f(n) = g(n)$ for all $n$ so that $f$ and $g$ are the same function. On the other hand, if $\mu$ is Lebesgue measure restricted to $\mathbb N$, the fact that $\mu(\mathbb N) = 0$ implies that $\mu(\{n \in \mathbb N \mid f(n) \not= g(n)\}) = 0$ for any two functions $f$ and $g$, so that any two functions $f$ and $g$ are equal a.e. Different measures yield totally different results.

  • $\begingroup$ Yes yes, I'm aware of the above. I should have mentioned that this is just the Lebesgue Measure. So, it is true then, that we may say a.e. even if the property is only defined on a set of measure 0 w.r.t that particular measure? $\endgroup$ – Anthony Peter Oct 2 '15 at 20:37
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    $\begingroup$ No. You don't just "need a measure". The whole notion of "a.e." has to do with functions defined on a measure space. If $(X,\mathcal A,\mu)$ is a measure space then you talk about $f=g$ a.e. (wrt $\mu$) if $f$ and $g$ both have domain $X$ and they are equal except for a set of measure $0$. $\endgroup$ – David C. Ullrich Oct 2 '15 at 20:39
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    $\begingroup$ If you are restricting Lebesgue measure to $\mathbb N$, then every property holds almost everywhere on $\mathbb N$. It is true that $n = n+1$ a.e.! $\endgroup$ – Umberto P. Oct 2 '15 at 20:40
  • $\begingroup$ @Umberto P. Well that's unfortunate... so for this particular homework problem I'm assuming it was meant for the functions to be defined on a set of non-zero measure. $\endgroup$ – Anthony Peter Oct 2 '15 at 21:03

Here's an example to show that your concern about isolated points is justified. Let $f(x)=0$ for all $x\in{\Bbb R}$ and $g(x) = \max(x,0)$. These two functions are continuous and the are equal at (Lebesgue) a.e. point of $B:=(-\infty,0]\cup\{1\}$. But they are not identically equal on $B$.


If you have two functions $f$, $g$ that are defined and continuous on $\mathbb{R}$, then $f=g$ if the two agree a.e. with respect to Lebesgue measure. That may be how the question was meant to be interpreted, because that would be the default for a.e. on $\mathbb{R}$ if no measure is mentioned.

Assuming this interpretation: If the exceptional set is $E$, then you only have to note that $(x-\delta,x+\delta)$ contains a point of $\mathbb{R}\setminus E$ for every $\delta > 0$, which is enough to force $f(x)=g(x)$ for an arbitrary $x$ because $f$ and $g$ are continuous.


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