# Measure, Integration & Real Analysis Sheldon Axler Section 2B Exercise 16

I apologize for the title in advance, but this is a long problem. So, here it is:

Suppose $$\mathcal{S}$$ is a $$\sigma$$-algebra on a set $$X$$ and $$A \subset X$$. Let $$\mathcal{S}_A = \{E \in \mathcal{S} : A \subset E \text{ or } A \cap E = \emptyset\}$$

In the first question in this exercise, we're asked to show that $$\mathcal{S}_A$$ is a $$\sigma$$-algebra, which I have done and we can certainly use it if necessary in the second question, which is the one I'm struggling with. Here is the entire question:

Suppose $$f: X \rightarrow \mathbb{R}$$ is a function. Prove that $$f$$ is measurable with respect to $$\mathcal{S}_A$$ if and only if $$f$$ is measurable with respect to $$\mathcal{S}$$ and $$f$$ is constant on $$A$$

I was able to show that if $$f$$ is measurable with respect to $$\mathcal{S}_A$$ then it would be measurable with respect to $$\mathcal{S}$$ simply because $$f^{-1}(B) \in \mathcal{S}_A$$ for any Borel set $$B$$, but by the way $$\mathcal{S}_A$$ is defined, that would mean $$f^{-1}(B) \in \mathcal{S}$$ as well. The part that I'm struggling with is to first show that not only $$f$$ would be measurable with respect to $$\mathcal{S}$$ but also $$f$$ would be constant on $$A$$, and second is to show the converse to conclude the equivalence that is required. So, would anyone be able to help, please?

To show that $$f$$ is a constant on $$A$$ take $$a,b \in A$$ and suppose $$f(a)=c \neq d=f(b)$$. Then $$f^{-1} (\{c\})$$ and $$f^{-1} (\{d\})$$ belong to $$\mathcal S _A$$ but $$A$$ is neither contained in $$f^{-1} (\{c\})$$ nor is it disjoint from it. [ $$b \in A$$ and $$b \notin f^{-1} (\{c\})$$; Also, $$a \in f^{-1} (\{c\}) \cap A$$].

This contradiction proves that we must have $$c=d$$ or $$f(a)=f(b)$$.

The converse part is really trivial. If $$f$$ is a constant on $$A$$ what is the inverse image of any Borel set?

• The inverse image of any Borel set would be contained in $\mathcal{S}$, but how would I show that either $A \subset f^{-1}(B)$ or $A \cap f^{-1}(B) = \emptyset$, so that I can conclude that $f$ is measurable with respect to $\mathcal{S}_A$ as well.
– user831321
Feb 26, 2021 at 6:08
• @hdmovies598 $A \cap f^{-1}(B)$ is the inverse image of $B$ under the restriction of $f$ to the set $A$. Since this restriction is a constant this set is either empty or equal to $A$. [If you are still confused take a constant function on some set $Y$ and show that the inverse image of any set can only be $Y$ or empty]. Feb 26, 2021 at 6:11
• I see now, that makes sense, thanks for your explanation.
– user831321
Feb 26, 2021 at 6:14