Find a suitable function $\alpha: [0,2] \, \to \mathbb{R}$ such that $\int_{0}^{2}fd\alpha = f(0) + f(1) + f(2)$ for any continuous $f$ on $[0,2]$.
I was thinking we can make $\alpha(x) = -1$ if $x = 0$ and $1$ if $x = 2$ then $0$ otherwise, but how can we add in the $f(1)$ to that?
Let $\delta(x)$ be the dirac delta function, i.e. $\delta(x)=\begin{cases}0 & x\neq 0,\\ \infty & x=0\end{cases}$ and let $H(x)$ be the unit step function, i.e. $H(x)=\begin{cases}0 & x<0,\\ 1 & x\geq 0\end{cases}$.
Note that,
$$\int\limits_{0}^2 f(x)(\delta(x) + \delta(x-1)+\delta(x-2))dx=f(0)+f(1)+f(2)$$
So, we would really like for,
$$d\alpha = (\delta(x) + \delta(x-1)+\delta(x-2))dx$$
With this in mind, we have,
$$\frac{d\alpha}{dx}=\delta(x)+\delta(x-1)+\delta(x-2)$$
Integrating w.r.t. $x$ gives the answer,
$$\alpha(x) = H(x) + H(x-1) + H(x-2)$$