Let $a,b\in\mathbb R$ with $a<b$ and $f:[a,b]\to\mathbb R$. Assume that there exists a Riemann integrable function $g:[a,b]\to\mathbb R$ such that $f=g$ almost everywhere.
Then we can NOT conclude that $f$ is Riemann integrable. Indeed, $f$ can be unbounded: taking $a=0, b=1$, $g\equiv0$ and
$$f(x)=\begin{cases}\frac1x,& x\in(0,1]\cap\mathbb Q,\\ 0, & \text{otherwise.}\end{cases}\tag{$*$}\label{star}$$
However, if we also know that $f$ is a derivative, i.e., if there exists a differentiable function $F:[a, b]\to\mathbb R$ such that $f=F'$, then can we conclude that $f$ is Riemann integrable?
According to On the Riemannian integrability of the bounded derivative (now we don't assume the boundedness of $f$), it suffices to show that $f:[a,b]\to\mathbb R$ is bounded under the stronger condition that $f$ is a derivative.
Note that the function $f$ given by \eqref{star} is not a derivative, since it does NOT have the intermediate value property (see Darboux's theorem).
Best Answer
Since $f$ is equal a.e. to $g$, $f$ is Lebesgue integrable. Now for $a < x < y < b$, $$F(y) - F(x) = \int_x^y f(t)\; dt = \int_x^y g(t)\; dt$$ (see e.g. Rudin, Real and Complex Analysis, Theorem 8.21). Thus $$(y-x) \inf_{[x,y]} g \le F(y) - F(x) \le (y - x) \sup_{[x,y]} g $$ and this implies that if $y \in [x,z]$ with $x < z$, $$ \inf_{[x,z]} g \le f(y) \le \sup_{[x,z]} g$$
Consider a partition $\mathcal P$ of $[a,b]$ by $a = x_0 < x_1 < \ldots < x_n = b$ and the corresponding upper and lower sums $U(g;\mathcal P) = \sum_{i=1}^n (x_i - x_{i-1}) \sup_{[x_{i-1},x_i]} g$ and $L(g;\mathcal P) = \sum_{i=1}^n (x_i - x_{i-1}) \sup_{[x_{i-1},x_i]} g$. We have $$L(g;\mathcal P) \le L(f, \mathcal P) \le U(f, \mathcal P) \le U(g, \mathcal P)$$ so the Riemann integrability of $g$ implies that of $f$.