Consider $f_\lambda: \Bbb{R}_+ \to \Bbb{R}_+$,$$f_\lambda(t) = e^{-\lambda t}$$
Now consider the finite linear combinations of these functions (exponential polynomials)
$$ S(t) = \sum_{i = 1}^N \alpha_i f_{\lambda_i}(t)$$
The question is: given $x>0$, can we find ${S_n}$ such that $\int_0^\infty S_n (t) g(t)\, dt \to g(x)$? That is, $S_n(t)$ is some approximation of the Dirac measure in $x$.
Attempt: Let $\phi (t)\in C^{\infty}_c$ be a smooth function with compact support such that $\int_{-1}^1 \phi (t)\, dt = 1$. By the Stone-Weierstrass theorem, since $\mathcal{S}= \{\text{ exponential polynomial }\}$ separates points, has the constant function $1$. We see that for any $n$, $\phi_n(t) = n\phi( n(t - x))$, there is a $S_n \in \mathcal{S}$ such that $$\sup_{0 \leq t \leq x + n}\|\phi_n - S_n \|\leq \frac{1}{n^2} $$
$$S_n = \sum_{i = 1}^{N(n)} \alpha^n_i f_{\lambda^n_i}(t) $$ Moreover, since $\int_T^\infty e^{-\lambda t}\, dt = \frac{e^{-\lambda T}}{\lambda}$ it seems that we could somehow compute
$$\int_0^\infty S_n(t)g(t)\, dt = g(x) + \int_0^\infty (S_n(t) - \phi_n(t))g(t)\, dt + \int_0^\infty \phi_n(t)(g(t) -g(x)) \, dt $$
The error term $$ e_n = \int_0^\infty (S_n(t) - \phi_n(t))g(t)\, dt + \int_0^\infty \phi_n(t)(g(t) -g(x)) \, dt$$ can be divided into $$\int_0^\infty \phi_n(t)(g(t) -g(x)) \, dt \to 0 $$ and $$ \int_0^\infty (S_n(t) - \phi_n(t))g(t)\, dt = \int_0^{x+ n} (S_n(t) - \phi_n(t))g(t)\, dt + \int_{x+n}^\infty (S_n(t) - \phi_n(t))g(t)\, dt\\ \leq \frac{1}{n^2} \|g\| + \sum_{i = 1}^{N(n)} |\alpha_i^n| \frac{e^{-\lambda^n_i T}}{\lambda^n_i}\|g\|\ $$
But since $\alpha_i^n$ and $\lambda_i^n$ change with $S_n$ I am having trouble with the last bound.