What kind of proof is this? - Examining all the possibilities.

Question

To prove the following : If $f : X \rightarrow Y$ is a measurable function, and $E$ is a Borel set, then $f^{-1} (E)$ is a measurable set.

Prove) First, define $\Omega$ a collection of all $E \subset Y$ such that $f^{-1} (E)$ is measurable. Then we easily see that $\Omega$ is $\sigma$-algebra in $Y$. Since the measurablity of $f$, $\Omega$ contains all the open sets, and then $\Omega$ contains all Borel sets in $Y$. Then we are done.

This proof is unfamiliar to me. I could fully understand the context but I cannot apply this to the other. Is there anyone who can give any comment to me on this?

Answer 1

This is a standard kind of proof that is used in situations where we have inductively defined collections.

In this case, the inductively defined collection is the class of Borel sets. The Borel sets on a space are the smallest class of subsets of that space which contains all the open sets and is closed under countable unions and under complements. The thing that makes this an inductive definition is that it contains some "basic" elements (the open sets), some rules/methods for making more elements (taking countable unions, and taking intersections), and the clause that the class being defined is the smallest class which contains the basic elements and is closed under the specified rules/methods.

The way that the proof in the question works is this. It considers the class $\Omega = \{ E \subseteq Y : f^{-1}(E) \text{ is measurable}\}$. The goal is to show that all Borel sets are in $\Omega$. Perhaps $\Omega$ will contain more sets, but that isn't relevant to the proof.

To show that $\Omega$ contains all the Borel sets, we only need to show that $\Omega$ itself contains all the open sets, and that $\Omega$ is closed under taking countable unions and taking complements. This is because the Borel sets are the smallest such class, so if $\Omega$ has those properties then all the Borel sets have to be included in $\Omega$.

Another setting where proofs like this are used is with the natural numbers: in that setting this proof method is just the principle of mathematical induction. The natural numbers are the smallest set of numbers containing $0$ and closed under the function $s(x) = x+1$. This is an inductive definition, so we can use the same proof method with it. To show that some set $L$ contains all the natural numbers, it is therefore sufficient to prove that $0 \in L$ and that $L$ is closed under the function $s$. Then, because the natural numbers are the smallest set with those properties, the natural numbers must be a subset of $L$.

Answer 2

I think with "We easily see" and "then we are done" you haven't completed the proof. These aren't statements which could be written down in logic to deduce the conclusion.

By taking more care to write a series of propositions, only arriving at any one if it can be deduced from prior propositions, and then ask yourself again, what type of proof is this, it better reveals the nature of the proof. In this case it appears to be a proof by deduction.

In this instance it appears there is some set of conditions which, if $f^{-1}(X)$ satisfies, are sufficient for it to be a measurable set. So we write out the statement $f^{-1}(E)$ is a measurable set if $x$ is true.

Then in a series of propositions we combine the facts we know to arrive at a set of true statements about $f^{-1}(E)$. If these conditions include $x$ then we have proven our conjecture by deduction.