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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

1. $\mu(\emptyset)=0$,
2. 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.)

My attempt (which is wrong in general, see below): I think yes. I would proceed as follows:
First, prove that there exists a unique signed measure $\mu$ such that $\mu(]a, b])=f(b)-f(a)$ for all reals $a\le b$.
Second, show that this $\mu$ satisfies \eqref{*}.

Does this work? And is there some easy well-known way to prove these? It seems that \eqref{*} is almost like integration-by-parts for Riemann-Stieltjes integrals.

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

1. $\mu(\emptyset)=0$,
2. 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.)

Edit. This question suffers from subtleties that arise out of signed measures, see the answer(s) below...

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.)

My attempt: I think yes. I would proceed as follows:
First, prove that there exists a unique signed measure $\mu$ such that $\mu(]a, b])=f(b)-f(a)$ for all reals $a\le b$.
Second, show that this $\mu$ satisfies \eqref{*}.

Does this work? And is there some easy well-known way to prove these? It seems that \eqref{*} is almost like integration-by-parts for Riemann-Stieltjes integrals.

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

1. $\mu(\emptyset)=0$,
2. 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.)

My attempt (which is wrong in general, see below): I think yes. I would proceed as follows:
First, prove that there exists a unique signed measure $\mu$ such that $\mu(]a, b])=f(b)-f(a)$ for all reals $a\le b$.
Second, show that this $\mu$ satisfies \eqref{*}.

Does this work? And is there some easy well-known way to prove these? It seems that \eqref{*} is almost like integration-by-parts for Riemann-Stieltjes integrals.

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.)

My attempt: I think yes. I would proceed as follows:
First, prove that there exists a unique signed measure $\mu$ such that $\mu(]a, b])=f(b)-f(a)$ for all reals $a\le b$.
Second, show that this $\mu$ satisfies \eqref{*}.

Does this work? And is there some easy well-known way to prove these? It seems that \eqref{*} is almost like integration-by-parts for Riemann-Stieltjes integrals.

Edit. This question suffers from subtleties that arise out of signed measures, see the answer(s) below...

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.)

My attempt: I think yes. I would proceed as follows:
First, prove that there exists a unique signed measure $\mu$ such that $\mu(]a, b])=f(b)-f(a)$ for all reals $a\le b$.
Second, show that this $\mu$ satisfies \eqref{*}.

Does this work? And is there some easy well-known way to prove these? It seems that \eqref{*} is almost like integration-by-parts for Riemann-Stieltjes integrals.