Counter-Example for $f$ unbounded, $f$ continuous at $s$, and $\alpha(x)=I(x-s)$ such that $\int_a^b f \,d\alpha \neq f(s)$ (Rudin Thm 6.15)

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I'm reading Theorem 6.15 in Baby Rudin, which states that:

Theorem 6.15: If $a<s<b$, $f$ is bounded on $[a,b]$, $f$ is continuous at $s$, and $\alpha(x)=I(x-s)$, then $$\int_a^b f \,d\alpha = f(s)$$ where $I$ is the unit step function defined by $$I(x)= \begin{cases} 0, & \text{if $(x \leq 0)$} \\ 1, & \text{if $(x > 0)$ .} \\ \end{cases} $$

In case it is unclear, the integral above is a Riemann-Stieltjes integral.

I'm trying to think of an example where $f$ is unbounded on $[a,b]$ so that $\int_a^b f \,d\alpha \neq f(s)$, but I'm having a lot of trouble coming up with one.