Support of a positive measure

Question

Let $\mu$ be Lebesgue measure on $\mathbb{R}$ , $\mathcal{M}$ be Lebesgue $\sigma-$ algebra and $f\in C_{c}\left(\mathbb{R}\right)$ (continuous with compact support). Suppose $f\geq0$ over $\mathbb{R}$ and

$${\displaystyle \int_{\mathbb{R}}fd\mu=1}$$

Let

$$\nu\left(E\right)={\displaystyle \int_{E}}fd\mu\qquad\forall E\in\mathcal{M}$$

Assume that $\nu$ is a positive measure. Which set be support of $\nu$ ?

Thank you in advanced.

Answer 1

The support of the measure $\nu$,

$$\DeclareMathOperator \spt{spt} \spt \nu := \{x \in \mathbb{R}; \forall \delta>0: \nu(B(x,\delta))>0\},$$

equals the support of the function $f$.

Proof:

1. Let $x \in \{y \in \mathbb{R}; f(y) \neq 0\}=: U$. Since $f$ is continuous, we know that $U$ is open, i.e. there exists $\delta_0>0$ such that $B(x,\delta_0) \subseteq U$, i.e. $$f(y) > 0 \quad \text{for all} \, y \in B(x,\delta_0).$$ Using again that $f$ is continuous, we find $$\nu(B(x,\delta)) \geq \nu(B[x,\delta/2]) = \int_{B[x,\delta/2]} f(y) d\mu(y) \geq \delta \inf_{y \in B[x,\delta/2]} f(y)>0$$ for any $\delta \leq \delta_0$. Hence, $x \in \spt \nu$. As $\spt \nu$ is closed, we get $\spt f \subseteq \spt \nu$.
2. Let $x \in \spt \nu$. Suppose that $x \notin \spt f$. As $\spt f$ is closed (hence, its complement is open), we find $\delta>0$ such that $f|_{B(x,\delta)} = 0$. Obviously, this implies $\nu(B(x,\delta))=0$. Thus, by definition, $x \notin \spt \nu$. Contradiction.

Answer 2

I think the answer must be: $\text{supp}(\nu) = \text{supp} (f) \setminus \{\text{isolated points} \in \text{supp} (f)\}$. Or am I wrong something?