# Proving the measure of a set is between 0 and $2\epsilon$ using Countable Sub-additivity

Question: Let's consider a measure $\mu$: $B_{\mathbb{R}}$ $\rightarrow$ $[0,\infty]$ satisfying $\mu((a,b))$=$b-a$ for any $b>a$. Let {$x_j$}, where $j \in$ $\mathbb{N}$ be an enumeration of $\mathbb{Q}$ and let {$a_j$}, $j \in$ $\mathbb{N}$,be a sequence of positive numbers satisfying $\sum_{j=1}^{\infty} a_j< \epsilon$. Show that the set E:$\bigcup\limits_{j\in \mathbb{N}}^{} B(a_j,x_j)$, where ($B(a_j,x_j)$ means Ball with radius $a_j$ about $x_j$), is in $B_{\mathbb{R}}$ (Borel $\sigma$-algebra on ${\mathbb{R}}$) and that $0<\mu (E)<2 \epsilon$.

Proof (My attempt): Clearly, E is in $B_{\mathbb{R}}$. For all $j\in {\mathbb{N}}$, $B(a_j,x_j)$ is an open set. That is $B(a_j,x_j)$ $\in B_{\mathbb{R}}$ for all $j \in {\mathbb{N}}$. E is just the countable union of open sets and since $B_{\mathbb{R}}$ is a sigma algebra and is closed under countable union, E $\in B_{\mathbb{R}}$. (Not sure if this part is exactly correct).

Now, for the second part I'm struggling with showing that $0<u(E)< 2\epsilon$. I feel like I should use countable sub-additivity. The measure of $B(a_j, x_j)$ is $2a_j$ for all $j$ as each Ball has radius $a_j$. By countable sub-additivity: $\mu(\bigcup\limits_{j\in \mathbb{N}}^{} B(a_j,x_j))$$\leq$$\sum_{j=1}^{\infty}\mu(B(a_j,x_j))$=$\sum_{j=1}^{\infty} 2 a_j$=$2\sum_{j=1}^{\infty} a_j<2\epsilon$. (I'm not sure if this correct either). Now that I've shown the measure of E is less than $2\epsilon$, how do I show that it is greater than $0$? Or is that just implicit from the way the measure is defined?

• Measure is non-negative, yes. Your proof seems good. – Fimpellizieri Sep 3 '18 at 20:48
• Since $E\supset B(a_1,x_1)$, we have $\mu(E)\ge\mu(B(a_1,x_1)) = 2a_1 > 0$ as the $a_j$ are positive. – amsmath Sep 3 '18 at 20:50

Everything you've wondered about, you're correct about. Your proof that $E$ is in your sigma algebra is correct. Your proof that your set has measure at most $2\epsilon$ is correct. And you are correct in your guess that measure is at least positive. (You've sated it already: $\mu:B_\mathbb{R}\to [0,\infty]$)
• It has to be strictly positive but it is, e.g., because $\mu(E) \geq \mu(B(x_1,a_1))$ – ajr Sep 3 '18 at 20:50