Measure on Borel sets

Question

If $\mu_1,\mu_2$ are two Borel measures with $\mu_1([a,b])\le \mu_2([a,b])$ for all $a<b$, how we can prove that $\mu_1(A)\le \mu_2(A)$ for all Borel sets $A$? (All sets are from $\mathbb{R}$) Unfortunately, the set of intervals $[a,b]$ is not even a semi-ring, so I don't even know how to start here. Would be nice if someone has some ideas about it. Thanks in advance.

Answer 1

Disclaimer: This answer is much longer than a solution to this problem needs to be because I'm explaining everything in detail -- you can easily write this down in 2-3 lines.

I don't know if you're familiar with the Monotone Class Lemma, but this seems like a standard application. Anything that relates the properties of measures that hold under limits (in this case, weak inequality) is usually solvable using this lemma. So, for future reference, if you're up against a problem that asks gives you a property of measures that holds for a certain class of "nice" sets (in this case, intervals), and th eproperty holds under limits (in this case, the inequality), then MCL is probably a useful tool.

Step 1. Notice that this holds for half open half closed intervals of the form $(a,b]$. Now that $(a,b] = \bigcup_{n \in \mathbb{N}} [a+\frac{1}{n}, b]$ so, $$ \mu_{1}((a,b]) = \lim_{n \to \infty} \mu_{1}([a + \frac{1}{n}, b]) \leq lim_{n \to \infty} \mu_{2}([a+\frac{1}{n}, b]) = \mu_{2}((a,b]) $$

Step 2. Recall the Monotone Class Lemma (or look it up, if you aren't familiar with it): If $\mathcal{A}$ is an algebra then the $\sigma$-algebra generated by $\mathcal{A}$, denoted $\sigma(\mathcal{A})$, is equal to the smallest monotone class containing $\mathcal{A}$, denoted $\mathcal{M}(\mathcal{A})$. A monotone class is a collection of sets which is closed under increasing unions and decreasing intersections.

Step 3. We will now apply this result. Our algebra $\mathcal{A}$ will be the algebra generated by half open half closed intervals $(a,b]$ (which, you may recall, consists of disjoint unions of half-open half-closed intervals). If we can show that the set $$ B = \{A \in \mathcal{B}: \mu_{1}(A) \leq \mu_{2}(B)\} $$ is a monotone class which contains $\mathcal{A}$, then it must contain $\mathcal{M}(A)$, so by the Monotone Class Lemma is must contain $\sigma(\mathcal{A})$, i.e. all Borel sets. But this is fairly straightforward, since if $\{A_{n}\}_{n \in \mathbb{N}}$ is a collection in $B$ such that $A_{n} \uparrow A$, then $\mu_{1}(A) = \lim_{n \to \infty} \mu_{1}(A_{n}) \leq \lim_{n \to \infty} \mu_{2}(A_{n}) = \mu_{2}(A)$ (and same for decreasing sequences), which shows that $A \in B$. This shows that $B$ is a monotone class and we're done.