Why is $f$ measurable iff $f^{-1}\{(-\infty , c) \}$ for any $c\in\mathbb{R}$?

Question

I have just started a course in real analysis and I encountered this: Theorem 2.1.1 from Friedmans Modern Analysis, it is fairly straight forward modulo two technical details 1. and 2. below:

It begins with the following definition: Let $X_o$ be a measurable set in some fixed measurespace $(X,\mu,\sigma)$, then a function $f:X_o \rightarrow \mathbb{R}$ is measurable if the preimage of any open set is measurable.

The theorem states that for such a function; $f$ is measurable iff for all $c\in\mathbb{R}$ the preimage $f^{-1}\{(-\infty,c)\}$ and the sets $f^{-1}(\pm \infty)$ are measurable.

Proof: $\Rightarrow$ is immediate from the definition. In the book, the proof of the converse is as follows: Assume $f$ satisfies the above confitions then:

1. $f^{-1}\{(-\infty,c]\} = \cap_nf^{-1}\{(-\infty, c + \frac{1}{n})\}$, the RHS is measurable as the intersection of measurable sets of the $\sigma$-algebra, hence

2. $f^{-1}\{(a,b)\} = f^{-1}\{(-\infty,b)\} - f^{-1}\{(-\infty,a]\} $.

I understand that as an interval on the real line $(-\infty,c] = \cap_n (-\infty,c+\frac{1}{n})$, moreover I understand that as a set on the real line $(a,b) = (-\infty, b) - (-\infty, a]$ but how do I know from the definition of a measurable function that the two measurable sets in 1. and 2. are equal in the $\sigma$-algebra?

Perhaps it is just trivial and I can't see it, any insight to what I fail to understand would be helpful!

Answer 1

The correct name of the book you quote in your post is Foundations of Modern Analysis. The author is Friedman. (There is no "s" is the author's name.) The theorem you quote is not complete. Here is the one in the book:

Theorem 2.1.1 Let $f$ be a extended real-valued function defined on a measurable set $X_0\subset X$, $X$ a measure space. $f$ is measurable if and only if

• (i) for every real number $c$, the set $f^{-1}((-\infty, c))$ is measurable.
• (ii) the sets $f^{-1}(-\infty)$ and $f^{-1}(\infty)$ are measurable.

A key property of the inverse map $f^{-1}$ you need to know in order to understand the proof is that it commutes with unions, intersections and complements: $$ \begin{align} f^{-1}(\cup_{\alpha\in A} E_\alpha)&=\cup_{\alpha\in A} f^{-1}(E_{\alpha}),\tag{1}\\ f^{-1}(\cap_{\alpha\in A} E_{\alpha})&=\cap_{\alpha\in A} f^{-1}(E_\alpha),\tag{2}\\ f^{-1}(E^c)&=\big(f^{-1}(E)\big)^c.\tag{3} \end{align} $$

Note that this true for any function $f$. It has nothing to do with its measurability.

This property tells you why you have the two identities in your 1 and 2. For instance, (2) tells you that $$ f^{-1}((\infty,c])=f^{-1}(\cap_{n=1}^\infty A_n) = \cap_{n=1}^\infty f^{-1}(A_n) $$ where $A_n=(-\infty,c+\frac1n)$.

Answer 2

The collection of all open rays generates the Borel $\sigma$-algebra (i.e. the canonical $\sigma$-algebra over $\mathbb R$). By some theorem, it suffices to check measurability of a function on the generating set — in this case the collection of all open rays. Hence, once $f^{-1}\{(-\infty, c)\}\in\sigma$ for arbitrary $c\in\mathbb R$, $f$ is measurable. The theorem is, unsurprisingly, also (implicitly) used in the "$\Leftarrow$" direction: Since $f^{-1}\{(-\infty,c)\}\in\sigma$ for arbitrary $c\in\mathbb R$ it can be shown that this implies $f^{-1}\{(a,b)\}\in\sigma$. But the collection of open intervals also generates the Borel $\sigma$-algebra, hence we have proved measurability on a generator which implies measurability of the function.