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?
 A: 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.
