Show that $f$ is Riemann integrable and that $\int_a^b f (x)dx = 0$ Let $k \in \mathbb{R}$, $x_0 \in [a, b]$, and define $f:[a, b] \to \mathbb{R}$ via
$$
\begin{cases}
    0 & x \ne x_0\\
    k & x= x_0
\end{cases}
$$
Show that $f$ is  Riemann integrable and that $\int_a^b f (x)dx = 0$.
I know that I need to show that the lower and upper Riemann sums must be equal, but in my mind they are not equal, because at the point $x=x_0$, it will have an upper sum of $k$ but a lower sum of $0$. However, I understand and can visualize how the entire integral is equal to $0$.
Edit: I find the upper Riemann sum by
$$
U(f,P_n)=\sum_{i=1}^n M_i\Delta x_i.
$$
I know that $\Delta x_i =\frac{1}{n}$, but I don't know how to write $M_i$ because the function will suddenly jump up to $k$ at a certain point.

The upper and lower sums will not be equal, but the upper and lower integrals will be. I'm not sure what notation you use, so I will use the one I am most familiar with. For a partition $P=\{x_0,x_1,\ldots,x_n\}$, the upper and lower sums are respectively \begin{align*}
U(f,P) &= \sum^n_{i=1} \left(\sup_{x\in[x_{i-1},x_i]} f(x)\right)(x_i-x_{i-1}), \\
L(f,P) &= \sum^n_{i=1} \left(\inf_{x\in[x_{i-1},x_i]} f(x)\right)(x_i-x_{i-1}).
\end{align*} Note that the width of the intervals come into play here. You are right that the supremum or infimum of $f$ will take the value $k$ for one (or maybe two) of those intervals, but the width of those intervals could be very small so that contribution can be made as small as you like.
