Borel measure $\mu$ with $f'(a)$ an integral of $f$ w.r.t $\mu$ I am learning measure theory, and I noticed that if we integrate a function $f : \mathbb{R} \to \mathbb{R}$  with respect to a dirac measure at $a \in \mathbb{R}$, call it $\mu$, then
$$ \int f d\mu = f(a)$$
I am wondering how this generalizes, in particular I was wondering if there exits a borel measure (Possibly signed) $\mu$ such that
$$f'(a) = \int f d\mu$$
for $f \in C^\infty(\mathbb{R}) $
Thank you in advance!
My thoughts:  I am honestly a little clueless, I am learning from Rudin's RCA and I was thinking that maybe we could use the Riez representation theorem, maybe if we instead take $f$ so that $f \in C^\infty(\mathbb{R}) $ and it is compactly supported we can use it, but I am not sure where the linear functional would come from.
 A: This is impossible for a simple reason. We can have a measurable function $f$ which is nonnegative everywhere but where $f’(a) < 0$. In this case, we have $f’(a) < 0 \leq \int f d\mu$.
Edit: this is also impossible for a signed measure. In particular, consider a countable basis $(B_n)_{n \in \mathbb{N}}$ of open neighbourhoods of $a$, and consider any Borel set $S$ such that $a \notin S$. WLOG, suppose $B$ is a decreasing sequence; that is, $B_{n + 1} \subseteq B_n$ for all $n$. Then $S = \bigcup\limits_{n \in \mathbb{N}} S \setminus B_n$.
I claim that for all $n$, $\mu(S \setminus B_n) = 0$. This is because $\mu(S \setminus B_n) = \int \chi_{S \setminus B_n} d\mu = \chi_{S \setminus B_n}’(a) = 0$.
Thus, we can show that $\mu(S) = \lim\limits_{n \to \infty} \mu(S \setminus B_n) = 0$.
In slightly more generality, then, for all Borel sets $R$, we have
$$\mu(R) = \begin{cases}
  0 & a \notin R \\
  \mu(\{a\}) & a \in R
\end{cases}$$
since if $a \in R$, then $R = (R \setminus \{a\}) \cup \{a\}$.
In this case, we see that $\mu$ is a constant multiple of type Dirac delta, and hence does not have the sought-after property.
