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I am just starting learning about measure theory (what a great way to spend Christmas...!), and I am unclear on this following claim: Characterisation of Measurable Functions. It appears to be a very basic (and fundamental) result, but I'm not really sure how to show it. I'm only really interested in "$(2)$ and $(2')$". I'm sure that I can deduce the others from it.

This is what I've been thinking: Given measurable $f:E \rightarrow \Bbb R$, write $f = \lim f_n$, where $f_n$ is a simple function, which is known to be measurable. Then $f(x) \le \alpha$ just when when $x$ is in one of the sets in the simple function which has constant $\le \alpha$; eg $$f_n = \sum_{k=1}^{K_n} c_k^{(n)} 1(A_k^{(n)})$$ where $1(\cdot)$ is the indicator function, satisfies $fn(x) \le \alpha$ just when $x \in A_k^{(n)}$ with $c_k^{(n)} \le \alpha$. [I'm not 100% sure how this translates into the limit...].

An explanation would be most appreciated! Thanks! :)

UPDATE

The definition for a measurable set that I have been given is just the elements of the $\sigma$-algebra. Now for a function:

Let $(E,\mathcal E)$ and $(G,\mathcal G)$ be measurable spaces. We say that a function $f:E \rightarrow G$ is measurable if $f^{-1}(A) \in \mathcal E$ for every $A \in \mathcal G$.

Using the definition given in a link on the page linked above, the main part of my question ($(1) \iff (2)$) is fairly trivial. I'm interested to know how the above (equivalent, I assume!) definition gives this result.

Sam OT
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(1) and (2) are the same since (2) is the definition of measurable in the reference above. So, we need only show $(2) \Longleftrightarrow (2')$.

Clearly (2) $\Rightarrow$ (2').

Let $L_t = \{x | f(x) \le t \}$.

Suppose (2') holds and $\beta \in \mathbb{R}$. Let $\alpha_n \in \mathbb{Q}$ be a decreasing sequence such that $\alpha_n \downarrow \beta$.

Since $L_{\alpha_n} \in \Sigma$ for all $n$, we have $\cap_n L_{\alpha_n} \in \Sigma$, and since $L_\beta = \cap_n L_{\alpha_n}$, we have $L_\beta \in \Sigma$. Hence (2) holds.

Some notes: When the range of a function is a topological space, there is an implied Borel $\sigma$-algebra generated by the open sets. The definition in the question defines measurability for a function, but often (especially when the range is $\mathbb{R}$) the range $\sigma$-algebra is implicitly taken to be the Borel $\sigma$-algebra.

Suppose ${\cal C}$ is a collection of subsets of $G$ and suppose $f^{-1}(C) \in {\cal E}$ for all $C \in {\cal C}$, then it is straightforward to show that $f^{-1}(C) \in {\cal E}$ for all $C \in \sigma({\cal C})$.

From this we see that if $f^{-1}((-\infty,\alpha]) \in {\cal E}$ for all $\alpha$, then since ${\cal B} = \sigma (\{(-\infty,\alpha]\}_\alpha)$ (with ${\cal B}$ being the Borel sets of $\mathbb{R}$), we see that $f$ is (Borel) measurable. (The other direction is trivial.)

(Also see Definition of a measurable function?.)

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copper.hat
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  • See update to question re definitions. I didn't actually check what the definition for a measurable function given on the wiki page was - it is different to what I've been given. Using their definition, as you've said, $(1)$ and $(2)$ are the same (by definition). Thanks for showing $(2) \iff (2')$ though (well, $\Rightarrow$ is trivial - thanks for the $\Leftarrow$ part!). – Sam OT Jan 01 '15 at 00:23
  • You need to add more restrictions. Is ${\cal G}$ the collection of Borel measurable subsets? You need to connect ${\cal G}$ with the $\sigma$-algebra containing sets of the form $(-\infty, \alpha]$. – copper.hat Jan 01 '15 at 01:37
  • Ah ok, good point. Is it correct that the definition using "$f(x) \le \alpha$" is for when $(G,\mathcal G) = \Bbb R, \mathcal B$ (the Borel measure on $\Bbb R$, ie generated by the standard Euclidean topology)? Would make a lot more sense otherwise what does $f(x) \le \alpha$ even mean? =P – Sam OT Jan 01 '15 at 10:58
  • There seems to be broadly two sorts of (related) definitions used, the first being the one you have above and the other, for real valued functions, being Borel measurable. In the latter case, it is easy to see that it corresponds to the statement in your previous comment. – copper.hat Jan 01 '15 at 18:19
  • So my main question really is, how can I show that they two definitions are equivalent (when they are both defined, so for real-valued functions being Borel measurable)? – Sam OT Jan 01 '15 at 18:54
  • I added some notes, hopefully these address your questions. – copper.hat Jan 03 '15 at 20:13
  • Thanks, I'll read that tomorrow morning - sleep now! =P (I'm in the UK.) – Sam OT Jan 04 '15 at 00:00
  • Just read it - that makes sense now, thanks. I guess it's one of those things where sometimes, instead of considering all sets, you can consider a generating set. Similar to sometimes when taking a sup/inf, you don't need to do all sets but just certain generators? – Sam OT Jan 06 '15 at 10:28
  • Part of the issue is that it is hard to explicitly characterise a Borel set. – copper.hat Jan 06 '15 at 17:18
  • It's the $\sigma$-algebra generated by the set of open sets in $\Bbb R$, which is generated by the set of intervals $(-\infty, \alpha)$. I guess characterising something generated like that isn't that easy... – Sam OT Jan 07 '15 at 13:09
  • Well, I was thinking more along the lines of http://en.wikipedia.org/wiki/Borel_hierarchy... – copper.hat Jan 07 '15 at 19:07