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.
 A: 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:


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*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 $\infty$?

*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 $b_n=\frac{1}{2}$, $\forall n<\omega$.
