# Working with this 'almost everywhere' statement

Consider a real function $$f$$ and take $$\mathbb R$$ together with Lebesgue measure $$m$$. I want to check if I have the right reasoning with regards to working with the following "almost everywhere" statement; I have that $$f(x)\neq0$$ for almost all $$x\in\mathbb R$$.

Now, I know that this means that the set of $$x\in\mathbb R$$ for which $$f(x)=0$$ is a null set with respect to Lebesgue measure. That is, $$m(\{x\in\mathbb R:f(x)=0\})=0$$. But since the statement that $$f(x)\neq0$$ holds for almost all $$x\in\mathbb R$$, does that mean that $$m(\{x\in\mathbb R:f(x)\neq0\})=m(\mathbb R)=+\infty$$? My thinking is that since the statement holds for all $$\mathbb R$$ except possibly on a subset of $$\mathbb R$$ with measure zero, then the measure of the set on which the statement holds must coincide with the measure of $$\mathbb R$$; is this the correct reasoning?

• Yes, you are right. – Kabo Murphy Nov 13 '18 at 8:26
• @mathworker21, I don't think that it is too worrisome (unless you can indicate why it ought to be). I was unsure how to make use of the additivity property of the measure in a rigorous manner, but I see now how designating this set as $E$ makes that much clearer. – Jeremy Jeffrey James Nov 13 '18 at 8:33
• @JeremyJeffreyJames my apologies for the comment. I was wrong – mathworker21 Nov 13 '18 at 8:45

No issues with what you have said : $$f(x) \neq 0$$ almost everywhere, means that the set $$\{x : f(x) = 0\}$$ has Lebesgue measure zero, which implies that the set $$\{x : f(x) \neq 0\}$$ has the same Lebesgue measure as $$\mathbb R$$, which is $$+\infty$$.
However, note that the converse is not true : if I tell you that the set $$\{x : f(x) \neq 0\}$$ has measure $$+\infty$$, then this does not imply that $$f(x) \neq 0$$ almost everywhere. For example, take $$f(x) = 0$$ if $$[x]$$ is a multiple of $$2$$, and $$1$$ otherwise. Then $$\{x : f(x) \neq 0\}$$ and $$\{x : f(x) = 0\}$$ have measure $$+\infty$$.
In short, I want to tell you that when you say "$$f(x) \neq 0$$ almost everywhere means $$\{x : f(x) \neq 0\}$$ has measure infinite", we say it with the idea that the left side of the "means" is like a definition and the right side is an equivalent statement. However, the "means" for you, is an implication : the statements are not equivalent, as I showed. You should therefore be careful about a definition and an implication : that $$\{f(x) =0\}$$ has Lebesgue measure zero is the definition of $$f(x) \neq 0$$ almmost everywhere. On the other hand, that $$\{f(x) \neq 0\}$$ has full measure is an implication of $$f(x) \neq 0$$ everywhere, and not equivalent to it. As long as you are aware of this, you can be relatively be relaxed about what you are writing.