Let $I\subset\mathbb{R}$ denote an open and bounded interval of the real line, $H\_0^1(I)$ all quadratic integrable Sobolev functions and $C(\bar{I})$ all continuous functions on said interval. 


Since the embedding $H\_0^1(I)\hookrightarrow C(\bar{I})$ holds, we know that the delta distribution (point evaluation) is a linear functional on $H\_0^1(I)$, i.e., for all $s\in I:\delta\_s\in H^{-1}(I)$ holds. So for every $s\in I$ and every $f\in H\_0^1(I)$ Riesz representation theorem guarantees the existance of a uniquely determined $g\in H^1\_0(I)$ such that  

$$
 \langle \delta\_s, f\rangle\_{H^{-1},H\_0^1} = \delta\_s(f) = f(s) = \int\_I g'\_s(t)\cdot f'(t)dt
$$
holds. My question: Is the function $(s,t)\mapsto g\_s(t)$ Lebesgue measureable in $I^2\subset\mathbb{R}^2$?



Measurability of specific function
It certainly is measurable. In fact, you may find an explicit formula for it.


If we take $I = (0,1)$, then $g\_s$ is simply given by $g\_s(t) = \operatorname{min}(s,t) - st$.


How did I find this? Well, at first you might think of trying to choose $g\_s$ so that $g\_s'(t) = 1\_{[0,s]}(t)$; then you'd have $\int\_0^1 g\_s'(t) f'(t)\,dt = \int\_0^s f'(t)\,dt = f(s) - f(0) = f(s)$ since $f \in H\_0^1((0,1))$ vanishes at $0$ and $1$. That would yield $g\_s(t) = \operatorname{min}(s,t)$. Unfortunately such $g\_s$ is not in $H\_0^1$ itself because $g\_s(1) = s \ne 0$. 


However, we also note that if we subtract any constant from $g\_s'$, so that $g\_s'(t) = 1\_{[0,s]}(t) - c$, then we still have $$\int\_0^1 g\_s'(t) f'(t) \,dt = \int\_0^s f'(t) - c\int\_0^1 f'(t) = (f(s) - f(0)) - c(f(1) - f(0)) = f(s).$$
This would yield $g\_s(t) = \operatorname{min}(s,t) - ct$. And now we see that choosing $c=s$ results in $g\_s(1) =0$.


