Assume $f : \mathbb R \to \mathbb R$ is integrable on the interval $I = [a, b]$ and let $c \in (a, b)$. Then define \begin{equation} g(x) = \left\{ \begin{aligned} f(x) &, x\neq c \\ y &, x = c\,. \end{aligned} \right. \end{equation}
Is $\int_a^b f(x) \,\mathrm d x = \int_a^b g(x) \,\mathrm d x$?
My thoughts
Since $f$ is integrable, then is $g$, as they differ only by a single value $y$ at one point $c$ in the interval $(a, b)$, and that value is implied to be finite. The integrability of both functions implies that they are bounded, meaning there exist $M_1, M_2 \in \mathbb R^+$, so that \begin{equation} M_1 > |f(x)| \quad\text{and}\quad M_2 > |g(x)| = \left\{ \begin{aligned} |f(x)| &, x \neq c \\ |y| &, x = c\,. \end{aligned} \right. \end{equation}
As both functions are bounded, we can also conclude, that at least for a certain $M \in \mathbb R^+$, \begin{equation} M > |f(x) - g(x)| = \left\{ \begin{aligned} 0 &, x \neq c \\ |f(x) - y| &, x = c\,. \end{aligned} \right. \end{equation}
Now, following the known rules of integration we can conclude that \begin{align} \left| \int_a^b f(x) - g(x)\, \mathrm d x \right| &= \left| \int_a^c f(x) - g(x)\, \mathrm d x + \int_c^b f(x)\,\,\mathrm d x \right| \\ &\leq \left| \int_a^c f(x) - g(x)\, \mathrm d x \right| + \left|\int_c^b f(x) - g(x)\,\,\mathrm d x \right| \\ &< M(c - a) + M(b - c)\,. \end{align}
Now all that is left to show is, that we can choose $M = \varepsilon > 0$, and we are done. But how do I do that?