Let $\Omega$ be an open subset of $\mathbb{R}^n$. If $\mu$ is a Radon measure on $\mathbb{R}^n$, then we denote with $T_\mu$ the distribution defined as follow: $$<T_\mu, \phi>=\int_\Omega \phi d\mu \quad \forall \phi \in C^\infty_c(\Omega).$$ In the same way, if $f\in L^1_{loc}(\Omega)$, then we denote with $T_f$ the distribution defined as follow: $$<T_f, \phi>=\int_\Omega \phi fdx \quad \forall \phi \in C^\infty_c(\Omega).$$

I should show that if $\mu$ and $f$ are, respectively, a Radon measure and a locally integrable function such that $T_\mu = T_f$, then $\mu << \mathcal{L}^n$. Please, can someone help me?

  • $\begingroup$ is $\mathcal{L}^n$ Lebesgue measure in $\mathbb{R}^n$? $\endgroup$
    – Ali Mezher
    Commented Feb 27 at 15:48

1 Answer 1


Let $\Omega$ be an open subset of $\mathbb{R}^n$. Suppose $\mu$ is a Radon measure on $\mathbb{R}^n$, and $f$ is a locally integrable function on $\Omega$, such that for the distributions $T_\mu$ and $T_f$ defined by

\begin{align*} \langle T_\mu, \phi \rangle &= \int_\Omega \phi \,d\mu, \quad \forall \phi \in C^\infty_c(\Omega), \\ \langle T_f, \phi \rangle &= \int_\Omega \phi f \,dx, \quad \forall \phi \in C^\infty_c(\Omega), \end{align*}

we have $T_\mu = T_f$. We aim to show that $\mu \ll \mathcal{L}^n$, where $\mathcal{L}^n$ is the Lebesgue measure on $\mathbb{R}^n$.

If $T_\mu = T_f$ for all $\phi \in C^\infty_c(\Omega)$, then $\mu \ll \mathcal{L}^n$.

The condition $T_\mu = T_f$ implies that for every $\phi \in C^\infty_c(\Omega)$, \begin{equation} \int_\Omega \phi \,d\mu = \int_\Omega \phi f \,dx. \end{equation}

To show $\mu \ll \mathcal{L}^n$, it suffices to demonstrate that for any measurable set $E \subseteq \Omega$ with $\mathcal{L}^n(E) = 0$, we have $\mu(E) = 0$.

Assume, for the sake of contradiction, that there exists a set $E \subseteq \Omega$ with $\mathcal{L}^n(E) = 0$ but $\mu(E) > 0$. Since $\mu$ is a Radon measure, it is inner regular, allowing us to find compact subsets $K \subseteq E$ with $\mu(K) > 0$.

Consider a sequence of smooth functions $\{\phi_n\} \subseteq C^\infty_c(\Omega)$ such that $\phi_n \rightarrow \chi_K$ pointwise almost everywhere, where $\chi_K$ is the characteristic function of $K$, and $0 \leq \phi_n \leq 1$ with $\text{supp}(\phi_n) \subseteq U$ for some open $U \subset \subset \Omega$ containing $K$.

Using the equivalence $T_\mu = T_f$, for these $\phi_n$, we get \begin{equation} \int_\Omega \phi_n \,d\mu = \int_\Omega \phi_n f \,dx. \end{equation} As $n \rightarrow \infty$, the dominated convergence theorem implies \begin{equation} \mu(K) = \int_K d\mu = \int_K f \,dx. \end{equation} However, since $\mathcal{L}^n(K) \leq \mathcal{L}^n(E) = 0$, and $f$ is locally integrable, the right-hand side is zero, which contradicts $\mu(K) > 0$.

Therefore, we conclude $\mu(E) = 0$ for any such $E$, implying $\mu \ll \mathcal{L}^n$.

  • 1
    $\begingroup$ Thank you very much for your answer $\endgroup$
    – Shiva
    Commented Feb 28 at 16:49

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