A question on Measure Theory (easy one)

Question

I think this needs a philosophical answer.

(1) It is know that every countably additive measure is finitely additive one, and

(2) every $\sigma$-algebra is an algebra.

BUT,

If one uses the Lebesgue measure on $[0,1]$, according to (1) this measure is finitely additive, but as a consequence, many open sets (not interval) in $[0,1]$ will be not measurable!

The question that I want it to be answered is:

Can we use countably additive measure by means of finitely additive one OR this means that when a countably additive measure satisfies a property, so is finitely additive one. What is the benefit on (1).

Please someone explain this, I am confused !

Answer 1

Definitoons vary, but the following should be fairly representative. Let $X$ be a set. A family $\mathcal{A}$ of subsets of $X$ is an algebra if

1. $X\in\mathcal{A}$ and $\emptyset\in\mathcal{A}$
2. Whenever $A\in\mathcal{A}$, then $X\backslash A\in\mathcal{A}$
3. Whenever $A_1\in\mathcal{A}$ and $A_2\in\mathcal{A}$,then $A_1\cup A_2\in\mathcal{A}$.

A family $\mathcal{A}$ of subsets of $X$ is a $\sigma$-algebra if

1. $X\in\mathcal{A}$ and $\emptyset\in\mathcal{A}$
2. Whenever $A\in\mathcal{A}$, then $X\backslash A\in\mathcal{A}$
3. Whenever $A_1,A_2,\ldots$ is a sequence of elements of $\mathcal{A}$, then $\bigcup_{n=1}^\infty A_n\in\mathcal{A}$.

The definitions only differ in condition 3. Now every $\sigma$-algebra is an algebra, for if $A_1\in\mathcal{A}$ and $A_2\in\mathcal{A}$, then by letting $A_n=\emptyset$ for $n\geq 3$, we get $$A_1\cup A_2=A_1\cup A_2\cup\emptyset\cup\emptyset\cup\ldots=\bigcup_{n=1}^\infty A_n\in\mathcal{A}.$$ It is slightly more difficult to see that an algebra may fail to be a $\sigma$-algebra.

A finitely additve measure is a function $\mu:\mathcal{A}\to\mathbb{R}\cup\{\infty\}$ with domain an algebra such that

1. $\mu(\emptyset)=0$.
2. $\mu(A)\geq 0$ for all $A\in\mathcal{A}$.
3. If $A_1\in\mathcal{A}$ and $A_2\in\mathcal{A}$ and $A_1\cap A_2=\emptyset$, then $\mu(A_1\cup A_2)=\mu(A_1)+\mu(A_2)$.

A measure is a function $\mu:\mathcal{A}\to\mathbb{R}\cup\{\infty\}$ with domain a $\sigma$-algebra such that

1. $\mu(\emptyset)=0$.
2. $\mu(A)\geq 0$ for all $A\in\mathcal{A}$.
3. If $A_n\in\mathcal{A}$ for all $n$ and $A_n \cap A_M=\emptyset$ whenever $m\neq n$, then $$\mu\left(\bigcup_{n=1}^\infty A_n\right)=\sum_{n=1}^\infty \mu(A_n).$$

Since every $\sigma$-algebra is an algebra, a measure is defined on an algebra. Now if $\mu$ is a measure on the $\sigma$-algebra $\mathcal{A}$ and $A_1\in\mathcal{A}$ and $A_2\in\mathcal{A}$ and $A_1\cap A_2=\emptyset$, then we can set $A_n=\emptyset$ for $n\geq 3$. Then $$\mu(A_1\cup A_2)=\mu(A_1\cup A_2\cup\emptyset\cup\emptyset\cup\ldots)=\mu\left(\bigcup_{n=1}^\infty A_n\right)$$ $$=\sum_{n=1}^\infty\mu(A_n)=\mu(A_1)+\mu(A_2)+0+0+\ldots=\mu(A_1)+\mu(A_2),$$ so every measure is also a finitely additive measure.

Remark: Definitions vary, and sometimes measures are defined even on algebras, in which case the definition of 3. has to hold only for the case that $\bigcup_{n=1}^\infty A_n\in\mathcal{A}$, which is automatic for $\sigma$-algebras. But by an important, if somewhat unintuitive, theorem, called the Caratheodory extension theorem, one can always enlarge in that case the measure to a $\sigma$-algebra.