MHB Show that the measure is equal to zero

[mathmari]

Hey!

Let $\mu$ be a Borel measure in $\mathbb{R}$ such that $\mu(I)\leq v^a(I)$ for each bounded interval $I$, where $a>1$.

Show that $\mu=0$.

Could you give some hints how to show this??

Do we maybe use the identity that for each rectangle R the outer measure of R is equal to the volume of R. But in this case we don`t have an outer measure... :/

 

[Euge]

What does $v^a$ stand for, mathmari?

 

[mathmari]

What does $v^a$ stand for, mathmari?

$v$ stands for volume.

 

[Euge]

So you're asking to show that if $\mu$ is a Borel measure on $\Bbb R$ such that for some $a > 1$, $\mu(I) \le [v(I)]^a$ for all bounded intervals $I$, then $\mu = 0$?

 

[mathmari]

So you're asking to show that if $\mu$ is a Borel measure on $\Bbb R$ such that for some $a > 1$, $\mu(I) \le [v(I)]^a$ for all bounded intervals $I$, then $\mu = 0$?

Yes! (Wondering)

 

[Euge]

Ok, let's start by showing $\mu(I) = 0$ for every bounded open interval $I$. Take an arbitrary, bounded open interval $I = (c,d)$. Given $n\in \Bbb N$, partition $I$ into $n$ disjoint intervals of length $(d - c)/n$:

$$\bigl(c, c + \tfrac{d - c}{n}\bigr), \bigl[c + \tfrac{d - c}{n}, c + \tfrac{2(d - c)}{n}),\ldots, \bigl[c + \tfrac{(n-1)(d - c)}{n}, d\bigr)$$

By hypothesis and countable additivity of $\mu$,

$$\mu(I) = \mu\bigl(c, c + \tfrac{d - c}{n}\bigr) + \mu\bigl[c + \tfrac{d - c}{n}, c + \tfrac{2(d - c)}{n}\bigr) + \cdots + \mu\bigl[c + \tfrac{(n-1)(d - c)}{n}, d\bigr)$$

$$ \le \Bigl(\frac{d - c}{n}\Bigr)^a + \Bigl(\frac{d - c}{n}\Bigr)^a + \cdots + \Bigl(\frac{d - c}{n}\Bigr)^a\; (\text{$n$ times})$$

$$= \frac{(d - c)^a}{n^{a-1}}.$$

Since $n$ is arbitrary and $a > 1$, taking the limit as $n\to \infty$ yields $\mu(I) \le 0$. Therefore $\mu(I) = 0$.

Next, let $I$ be any unbounded open interval. Then $I$ takes the form $(-\infty, x)$, $(x, \infty)$ (where $x\in \Bbb R$), or $(-\infty, \infty)$. In the first case, the sequence $\{(x - n, x)\}_{n = 1}^\infty$ increases to $(-\infty, x)$, and so $\mu(I) = \lim_{n\to \infty} \mu(x - n, x) = \lim_{n\to \infty} 0 = 0$. A similar argument applies to the second case. As for the last case,

$$\mu(I) = \mu\Bigl(\bigcup_{n = 1}^\infty (-n, n)\Bigr) \le \sum_{n = 1}^\infty \mu(-n,n) = \sum_{n = 1}^\infty 0 = 0,$$

hence $\mu(I) = 0$.

Now can you show that $\mu(G) = 0$ for every open set $G \subset \Bbb R$?