(T/F) If $\mu$ is a Borel measure on $\mathbb{R}$ and $A$ is a Borel set such that $\mu(A \cap K) = 0$ for all compact sets $K$, then $\mu(A) = 0$.

Question

I am trying to prove or disprove the following statement:

If $\mu$ is a Borel measure on $\mathbb{R}$ and $A$ is a Borel set such that $\mu(A \cap K) = 0$ for all compact sets $K$, then $\mu(A) = 0$.

I am looking for intuition behind the statement, since I am still trying to understand the Borel sigma-algebra.

Wikipedia indicates that some authors require that a Borel measure be locally compact. Do you require this, or have you defined a Borel measure as just one whose sigma algebra contains the Borel sigma algebra? — Sarvesh Ravichandran Iyer, Jan 28 '19 at 05:52
Our definition of a Borel measure is simply a measure on a Borel sigma-algebra. — , Jan 28 '19 at 05:54
Every $\sigma$-finite Borel measure on a Polish space is inner regular for compacta. — Henno Brandsma, Jan 28 '19 at 10:11

score 2 · Accepted Answer · 2019-01-28T06:10:42.780

2

$$ \mu(A)=\lim_{n\to\infty}\mu(A\cap[-n,n])=0. $$

edited Jan 28 '19 at 06:10

answered Jan 28 '19 at 06:02

Seems to be true, but what are you trying to show? How do you know that every compact set has the form $[-n,n]$ for some $n$? Isn't $[0,1]\cup[2,3]$ compact, for instance? – Jan 28 '19 at 06:09
https://math.stackexchange.com/questions/234292/continuity-from-below-and-above – Jan 28 '19 at 06:11
1

You stated that $\mu(A \cap K) = 0$ for all compact $K$. In particular, this is true for $K=[-n,n]$. – Jan 28 '19 at 06:14
Got it. Thank you. – Jan 28 '19 at 06:14

Mayuresh L · Answer 2 · 2019-01-28T07:17:43.277

0

Note by countable additivity of measure,

$0 \leq \mu(A)= \sum_{n\in \mathbb Z} \mu(A \cap (n,n+1]) \leq \sum_{n\in \mathbb Z} \mu(A \cap [n,n+1])= 0$

Hence $\mu(A)=0$.

edited Jan 28 '19 at 07:17

answered Jan 28 '19 at 06:19

Mayuresh L

2 Answers2