# Existence of a measure

I need help in showing that if $\alpha$ and $\beta$ are measures defined on $\mathfrak{A}$, and $\beta \leqslant\alpha$ then there is a measure $\lambda$ on $\mathfrak{A}$ such that $\lambda=\alpha-\beta.$

Attempt:

I started by defining $\lambda(E)=\alpha(E)-\beta(E)$. Then clearly, if $E=\emptyset$, then $\lambda(E)=0$. If $\beta(E)\lt \infty$, then $\lambda = \alpha- \beta$ is a measure. This is how far I've come.

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And what have you tried ? What definitions can you use to start you reasoning ? –  D. Thomine Jan 14 '12 at 22:14
What exactly does $\lambda = \alpha - \beta$ mean? If it means that $\lambda(E) = \alpha(E) - \beta(E)$ for all $E \in \mathfrak{A}$, then what should happen when $\alpha(E) = \beta(E) = \infty$? –  Nate Eldredge Jan 14 '12 at 22:25
@Nate: If $\alpha(E) = \beta(E) = \infty$, then $\lambda(E)$ is undefined... –  jojo Jan 14 '12 at 22:40
Is there a $\sigma$-finiteness hypothesis somewhere. Put it in the right place, and you can solve the problem. –  ncmathsadist Jan 14 '12 at 22:41
@ncmathsadist: No there isn't. Is it to be assumed? –  jojo Jan 14 '12 at 22:47

## 1 Answer

As D. Thomine noticed in his comments, the notion of $\alpha-\beta$, when $\alpha$ and $\beta$ are measures, is not clear. If it is simply $\alpha(E)-\beta(E)$ for $E\in\mathfrak{A}$, then there is problem which lays in infinite measures. For example, consider those two cases:

1. Let $\mathfrak{A}=\mathcal{P}(\mathbb{R})$. Let $\alpha(A)=0$ if $0\not\in A$, otherwise $\alpha(A)=\infty$. Similarly, let $\beta(A)=0$ if $1\not\in A$, otherwise $\beta(A)=\infty$. It is $\alpha(\{0,1\})=\beta(\{0,1\})=\infty$. What is the value of $\alpha(\{0,1\})-\beta(\{0,1\})$? $0$? Why? Why not $\infty$?

2. Let $\mathfrak{A}$ be a $\sigma$-algebra of Lebesgue measurable subsets of the real line $\mathbb{R}$. Let $\alpha$ be the Lebesgue measure and let $\beta=\frac{1}{2}\alpha$. It is $\alpha(\mathbb{R})=\beta(\mathbb{R})=\infty$, so what is the value of $\alpha(\mathbb{R})-\beta(\mathbb{R})$? $0$ or $\infty$? Or maybe any arbitrary $a\in\mathbb{R}$?

This shows that there is a problem with infinite measures. Fortunately, the situation in the second case can be rescued because of $\sigma$-finiteness of the Lebesgue measure.

Let us first work with the finite case, for this can be easily generalized to the $\sigma$-finite case.

Thus, let $\alpha$ be a finite measure. Since $\beta\le\alpha$, $\beta$ is also finite. We put $\lambda:\mathfrak{A}\to[0,\infty)$ as follows: $\lambda(E):=\alpha(E)-\beta(E)$ -- for each set $E\in\mathfrak{A}$ it is well-defined and non-negative, because $0\le\beta(E)\le\alpha(E)<\infty$. We have to check that $\lambda$ is a measure:

a) $\lambda(\emptyset)=\alpha(\emptyset)-\beta(\emptyset)=0-0=0$

b) let $\langle A_n\in\mathfrak{A}: n<\omega\rangle$ be a sequence of pairwise disjoint $\mathfrak{A}$-sets. We have: $$\lambda\left(\bigcup_{n<\omega}A_n\right)=\alpha\left(\bigcup_{n<\omega}A_n\right)-\beta\left(\bigcup_{n<\omega}A_n\right)=$$ $$=\sum_{n<\omega}\alpha(A_n)-\sum_{n<\omega}\beta(A_n)=\sum_{n<\omega}(\alpha(A_n)-\beta(A_n))=\sum_{n<\omega}\lambda(A_n)$$

The third transition is legal, because we work with convergent series (finiteness of $\alpha$ and $\beta$!). Thus we have proved that $\lambda$ is a measure.

Now, let us assume that $\alpha$ is a $\sigma$-finite measure defined on subsets of some space $X$. Again, since $\beta\le\alpha$, $\beta$ is also $\sigma$-finite. Let $X=\bigcup_{n<\omega}A_n$, where $A_n$ are measurable and of finite measure; wlog we can assume they are pairwise disjoint.

We define auxiliary functions $\lambda_n:\mathfrak{A}_n\to[0,\infty)$, $n<\omega$, where $\mathfrak{A}_n=\{E\cap A_n: E\in\mathfrak{A}\}$ (show that it is a $\sigma$-algebra on $A_n$!). Let $E\in\mathfrak{A}_n$. We put $\lambda_n(E)=\alpha(E)-\beta(E)$ ($=\alpha(E\cap A_n)-\beta(E\cap A_n)$). Analogously as previously, we can prove that $\lambda_n$ is a finite measure.

Now we can define $\lambda:\mathfrak{A}\to[0,\infty]$. It is done with the help of $\lambda_n$'s. Let $E\in\mathfrak{A}$, then let us put: $\lambda(E)=\sum_{n<\omega}\lambda_n(E\cap A_n)$.

For $E\in\mathfrak{A}$, we have: $\lambda(E)=\sum_{n<\omega}\lambda_n(E\cap A_n)=\sum_{n<\omega}(\alpha(E\cap A_n)-\beta(E\cap A_n))$. We cannot continue with this and write: $=\sum_{n<\omega}\alpha(E\cap A_n)-\sum_{n<\omega}\beta(E\cap A_n)=\alpha(E)-\beta(E)$ -- this is illegal. Compare it with a pair of series $\sum_{n<\omega}a_n, \sum_{n<\omega}b_n$, where $a_n=1$ and $b_n=1$, $\forall n<\omega$, or $a_n=1$ and $b_n=\frac{1}{2}$, $\forall n<\omega$.

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Is there another way of defining $\lambda$ other than $\lambda(E)=\alpha(E)-\beta(E)$? –  Colin Jan 16 '12 at 18:05
@Colin: But what do you want to obtain? We define $\lambda$ in such a way, because we want to make the intuition about subtraction of measures formal. So for finite measures it is clear. For $\sigma$-finite we have to use this partition of a space into smaller pieces on which measures are finite to avoid operations on infinity. –  Damian Sobota Jan 16 '12 at 19:01
Okay thanks. I'm bit confused about your last paragraph. where does the proof end? –  Colin Jan 16 '12 at 20:04
In the last but one paragraph, just with the definition of $\lambda$. The last paragraph does not belong to the proof -- it's a comment to it, and I did not prove that $\lambda$ is a measure -- I think it's clear or can be proved easily. The whole idea of the post is to approach as close as possible the intuition about subtracting measures. If you have any question concerning it, feel free to ask. –  Damian Sobota Jan 16 '12 at 21:21
Sorry to bother you again. why wont defining $\lambda(E)=\lambda_n(E\cap A_n)$ work? –  Colin Jan 16 '12 at 22:40