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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.

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  • $\begingroup$ For measures, there is no such thing as "the" support. For example, any measurable set containing $\text{supp}(f)$ is "a" support of $\nu$. $\endgroup$ – Stephen Montgomery-Smith Jun 14 '14 at 3:06
  • $\begingroup$ Ehm... I knew the definition of support of a function $f$ on a topological space (for example see Rudin Real and Complex Analysis) but I don't know the definition of support of a positive measure. Can you explain it? $\endgroup$ – chuyenvien94 Jun 14 '14 at 3:29
  • $\begingroup$ It is a measurable set $E$ such that $\nu(E \cap A) = \nu(A)$ for any measurable set $A$. Basically the set $E$ contains all the measure on $\nu$. $\endgroup$ – Stephen Montgomery-Smith Jun 14 '14 at 3:32
  • $\begingroup$ On the other hand, the web site en.wikipedia.org/wiki/Support_%28measure_theory%29 says I am wrong. $\endgroup$ – Stephen Montgomery-Smith Jun 14 '14 at 3:35
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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.
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  • $\begingroup$ Why we didn't use the assumption ${\displaystyle \int_{\mathbb{R}}fd\mu=1} $? $\endgroup$ – chuyenvien94 Jun 15 '14 at 0:48
  • $\begingroup$ I think $\spt\nu:=\overline{\{x\in\mathbb{R};\forall\delta>0:\nu(B(x,\delta))>0\}} $, that 's the closure of the set you specified. Because support of $f$ is the closure of the set of all $x$ such that $f(x) \neq 0$ $\endgroup$ – chuyenvien94 Jun 15 '14 at 1:02
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    $\begingroup$ @chuyenvien94 Note that the set $\{x; \forall \delta>0: \nu(B(x,\delta))>0\}$ is closed, i.e. $$\text{spt} \nu = \overline{\{x; \forall \delta: \nu(B(x,\delta))>0\}}.$$ Concerning $\int f \, d\mu = 1$: This condition ensures that $\spt \nu \neq \emptyset$. In fact, $\spt \nu \neq \emptyset \Leftrightarrow \int f \, d\mu >0.$$ $\endgroup$ – saz Jun 15 '14 at 7:07
  • $\begingroup$ +1. This is not the first question related to the support of a measure where the discussion is at risk of going astray hence a precise, rigorous, answer is all the more useful. $\endgroup$ – Did Jun 15 '14 at 8:15
  • $\begingroup$ @Did Thanks for your kind words. $\endgroup$ – saz Jun 15 '14 at 8:23
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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?

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