For this question, we can use a Darboux (or Riemann-Darboux) definition of the integral. There is a theorem which can be stated as follows:
Theorem: If $f$ is Darboux integrable on $[a,b]$ and $g$ is a bounded function such that $f(x) \neq g(x)$ at a finite number of points, then $g$ is also Darboux integrable.
An outline of this proof can be found here. However, if the condition of disagreement is changed to "the set of points at which $f(x) \neq g(x)$ has measure $0$," then this theorem seems to be false. That is, there seems to be functions f and g such that
$f$ is Darboux integrable on $[a,b]$ and $g$ is a bounded function on $[a,b]$ such that the set $E = \{x \in [a,b] : f(x) \neq g(x) \}$ has (Lebesgue) measure $0$, but $g$ is not Darboux integrable.
How could we find such a function? It seems to me that the only sets of measure $0$ are those sets that consist only of a finite number of points, so the two 'conditions of disagreement,' as I put it, would be equivalent, and no such pair of functions would exist. Even if the two ideas are not necessarily equivalent, they seem to at least be 'close enough' for this purpose. Nevertheless, by the definition of 'measure $0$,' it seems like they are different, and such functions do in fact exist. Perhaps I cannot see the example needed because I misunderstand what it means to have measure $0.$ So, to summarize, I think there is a counterexample to the following conjecture, and I would like help finding it:
Conjecture: If $f$ is Darboux integrable and $g$ is a bounded function on $[a,b]$ such that the set $E = \{ x \in [a,b] : f(x)\neq g(x) \}$ has measure $0$, then $g$ is also Darboux integrable.
Much appreciated.
Let $f=1$, $g=\chi_{{\bf{Q}}^{c}\cap[0,1]}$, then $f(x)\ne g(x)$ for all $x\in{\bf{Q}}\cap[0,1]$ and $g$ is bounded and not integrable.