# Distribution induced simultaneously by a local integrale function and a radon measure

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: $$=\int_\Omega \phi d\mu \quad \forall \phi \in C^\infty_c(\Omega).$$ In the same way, if $$f\in L^1_{loc}(\Omega)$$, then we denote with $$T_f$$ the distribution defined as follow: $$=\int_\Omega \phi fdx \quad \forall \phi \in C^\infty_c(\Omega).$$

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?

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

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 $$\phi \in C^\infty_c(\Omega)$$, $$$$\int_\Omega \phi \,d\mu = \int_\Omega \phi f \,dx.$$$$

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 $$$$\int_\Omega \phi_n \,d\mu = \int_\Omega \phi_n f \,dx.$$$$ As $$n \rightarrow \infty$$, the dominated convergence theorem implies $$$$\mu(K) = \int_K d\mu = \int_K f \,dx.$$$$ However, since $$\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 $$\mu \ll \mathcal{L}^n$$.

• Thank you very much for your answer Commented Feb 28 at 16:49