Theorem: A function $F$ is an indefinite integral if and only if it is absolutely continuous
Corollary: Every absolutely continuous function is the indefinite integral of its derivative. (thm 5.14,5.15 Royden,Real Analysis)
Since absolute continuity does not imply differentiability everywhere I am assuming that I could replace this with
"Every absolutely continuous function $F$ is the indefinite integral of any function $f$ which is equal to $F'$ almost everywhere",
and it is just assumed in analysis texts that the reader knows this?
This is of course technically lazy. However, there is a way it makes sense! The issue here is that technically, the function has a derivative - but said derivative is defined in a space where "almost everywhere" is enough. The derivative is not (quite) a function in the traditional sense - its value is undefined on a set of measure 0.
The way to justify this has to do with the theory of distributions and of the integrable function spaces $\mathcal{L}^p$. The derivative of any continuous (in fact, any locally integrable) function exists in a rigorous sense - it just may not be a function. In this case, what's being noted is that absolute continuity guarantees that the derivative of $f$ is "almost" a function, in the sense of being defined almost everywhere. In particular, this means it's well-defined in the space of integrable functions $\mathcal{L}^1$ - where functions are considered equivalent if they are equal almost everywhere.
Yes, this is a bit odd to wrap one's brain around - but it's a classic analyst's trick to "forget" for the moment that we're passing through a space in which functions are only defined up to changes on a set of measure 0.
EDIT: A quick clarification - what you state in the question, that $F$ is the integral of any function $f$ equal to $F'$ almost everywhere, is in fact ill-defined itself! After all, you're presuming that there's a function $F'$ to be equal to... you're passing through more general function spaces also, and just not noticing.