Edit. To clear up the confusion that I caused, I will define a signed measure here. The literature sometimes calls it "extended signed measure":
Definition. A signed measure $\mu$ on $(\mathbb R, \text{Borel sets})$ is a function $$\mu:\text{Borel sets}\to\mathbb R\cup\{\infty\}$$
or a function
$$\mu:\text{Borel sets}\to\mathbb R\cup\{-\infty\}$$
such that
- $\mu(\emptyset)=0$,
- for any disjoint Borel sets $A_1, A_2, A_3, \dots$, we have
$$\mu\left(\bigcup_{n\in\mathbb N} A_n\right)=\sum_{n\in\mathbb N} \mu(A_n),$$
with the convention that $\infty+\text{anything}=\infty$ and $-\infty+\text{anything}=-\infty$. Note that $\infty-\infty$ can never occur, since $\{-\infty, \infty\}\subset\operatorname{Image}\mu$ is impossible by definition.
Back to the question. Let $f\in L^1_{\text{loc}}(\mathbb R)$, i.e. $f$ is a locally absolutely integrable function. It is well-known that the distributional derivative of $f$ doesn't have to be expressible as a $L_{\text{loc}}^1$ function again. For example, if $f$ is the characteristic function of $[0,\infty[$ (or the characteristic function of $]0,\infty[$, for that matter), then its distributional derivative corresponds to the Dirac measure $\delta_0$, which has no $L^1_{\text{loc}}$-density with respect to the Lebesgue measure.
Similarly, if we have a measure on $\mathbb R$, its distributional derivative need not be a measure again. Continuing the above example, the distributional derivative of $\delta_0$ is given by the bounded linear operator
\begin{split}\delta_0': \mathcal C_{\text c}^\infty(\mathbb R) &\to \mathbb R \\ \phi&\mapsto-\phi'(0),\end{split}
which is not expressible as a measure on $\mathbb R$.
My question: Does every $L_{\text{loc}}^1$-function have a distributional derivative that can be expressed as a signed measure? More explicitly, if $f\in L_{\text{loc}}^1(\mathbb R)$, does there always exist a signed measure $\mu$ on $(\mathbb R, \text{Borel sets})$ such that
\begin{equation}\tag{*}\label{*}\bbox[15px,border:1px groove navy]{\int_{\mathbb R}\phi\,\mathrm d\mu = -\int_{\mathbb R}\phi'(t)\cdot f(t)\,\mathrm dt}\end{equation}
for every $\phi\in\mathcal C_{\text c}^\infty(\mathbb R)$ ? Note: In particular, I demand that $\int_{\mathbb R}\phi\,\mathrm d\mu$ is well-defined for every $\mathcal C_{\text c}^\infty(\mathbb R)$ (which, since $\mu$ is signed, can be actually quite a messy affair.)
Best Answer
If we understand signed Borel measure as a signed measure on the Borel sets such that $|\mu|(K)<\infty$ for every compact $K$, then there is actually a nice characterization all functions which have signed Borel as distributional derivatives:
First, every signed measure is a difference of two positive measures, so we may as well ask which functions have a positive measure as distributional derivative.
If $\mu$ is a positive Borel measure on $\mathbb R$, then $$ f\colon \mathbb R\to\mathbb R,\,f(x)=\begin{cases}-\mu([x,a))&\text{if }x<a\\ \mu((a,x])&\text{if }x\geq a\end{cases} $$ is locally integrable and it is not hard to check that it has weak derivative $\mu$. Since distributional derivatives are unique up to an additive constant, it follows that every $f\in L^1_{\mathrm{loc}}(\mathbb R)$ with $f'=\mu$ has an increasing representative. Conversely, if $f$ has an increasing representative, then $\langle f',\phi\rangle\geq 0$ for every $\phi\in C_c^\infty(\mathbb R)$ with $\phi\geq 0$. It is well-known that such distributions are represented by positive Borel measures.
Therefore $f\in L^1_{\mathrm{loc}}$ has a signed Borel measure as distributional derivative if and only if it has a representative that can be written as a difference of two increasing functions. Note that differences of monotone functions are exactly functions of locally bounded variation. So another way to phrase this result is to say that $f\in L^1_{\mathrm{loc}}$ has a signed Borel measure as distributional derivative if and only if it has a representative of locally bounded variation.