Consider $\mathbb{R}$ equipped with the Borel $\sigma$-field $\mathcal{B}(\mathbb{R})$ and the Lebesgue measure $\lambda$, i.e. $\lambda$ is the measure on $\mathcal{B}(\mathbb{R})$ which satisfies $\lambda((a,b))=b-a$ for $a,b\in\mathbb{R}$, $a.
Let $f:\mathbb{R}\to\mathbb{R}$ be any Borel measurable function and suppose we "change" $f$ in at most countably many points. If we call this new function $g$, then $f=g$ almost everywhere (with respect to $\lambda$). In particular, $f$ and $g$ have the same integral over any Borel set.
Let us formalize this: Let $(x_n)_{n\geq 1},(y_n)_{n\geq 1}\subseteq \mathbb{R}$ be two sequences of real numbers and define $ g(x)= \begin{cases} y_n\quad &\text{if } x=x_n,\\ f(x)&\text{if }x\neq x_n\text{ for all }n, \end{cases} $ i.e. we have "changed" $f$ along the sequence $(x_n)_{n\geq 1}$. Then $ \{x\in\mathbb{R}:f(x)\neq g(x)\}\subseteq \{x_n: n\geq 1\}, $ which has measure zero.