# Sufficient condition for measurability for functions with values in $[-\infty,\infty]$

I have proved the following result:

"Suppose $$(X,\mathcal{S})$$ is a measurable space and $$f:X\to [-\infty,\infty]$$ is a function such that $$f^{-1}((a,\infty])\in\mathcal{S}$$ for all $$a\in\mathbb{R}$$.

Then $$f$$ is an $$\mathcal{S}$$-measurable function."

and I would like to know if my proof is correct, thanks.

My proof (edited to take into account Alphie's answer):

Let $$\tau:=\{A\subset[-\infty,\infty]:f^{-1}(A)\in\mathcal{S}\}$$: if we can prove that the set of all Borel subsets of $$[-\infty,\infty]$$ is a subset of $$\tau$$ then the claim will follow because it will imply that $$f^{-1}(B)\in\mathcal{S}$$ for every Borel subset of $$[-\infty,\infty]$$, which is the definition of an $$\mathcal{S}$$-measurable function $$f:X\to [-\infty,\infty]$$.

We first prove that $$\tau$$ is a $$\sigma$$-algebra on $$[-\infty,\infty]$$:

• $$\emptyset\subset [-\infty,\infty],\ f^{-1}(\emptyset)=\emptyset$$ and being $$\mathcal{S}$$ a $$\sigma$$-algebra $$\emptyset\in\mathcal{S}$$ so $$\emptyset\in\tau$$;

• $$E\in\tau\Rightarrow E\subset [-\infty,\infty],\ f^{-1}(E)\in\mathcal{S}\Rightarrow X\setminus f^{-1}(E)\overset{(2.33a)}{=}f^{-1}([-\infty,\infty]\setminus E)\in\mathcal{S}$$ and since $$[-\infty,\infty]\setminus E\subset [-\infty,\infty]$$ we have $$[-\infty,\infty]\setminus E\in\mathcal{S}$$ so $$\tau$$ is closed under complementation;

• $$E_1, E_2,\dots\in\tau\Rightarrow f^{-1}(E_1),f^{-1}(E_2),\dots\in\mathcal{S}$$ so $$f^{-1}(\bigcup_{k=1}^{\infty}E_k)\overset{(2.33b)}{=}\bigcup_{k=1}^{\infty}f^{-1}(E_k)\in\mathcal{S}\Rightarrow \bigcup_{k=1}^{\infty}E_k\in\mathcal{S}$$ so $$\tau$$ is closed under countable unions.

So, $$\tau$$ is a $$\sigma$$-algebra on $$[-\infty,\infty]$$, as desired.

By hypothesis, $$\tau$$ contains $$\{(a,\infty]:a\in\mathbb{R}\}$$ and being a $$\sigma$$-algebra we have that it also contains $$\{[-\infty,b]:b\in\mathbb{R}\}$$, $$\{(a,b]:a,b\in\mathbb{R}\}$$ and thus all the intervals of the form $$(a,b)=\bigcup_{k=1}^{\infty}(a,b-\frac{1}{k}]$$, $$(-\infty, b)=\bigcup_{k=1}^{\infty} (-k,b-\frac{1}{k}]$$, $$[-\infty,b)=\bigcup_{k=1}^{\infty}[-\infty,b-\frac{1}{k}]$$ and $$[a,\infty]=[-\infty,\infty]\setminus [-\infty,a)$$, $$[-\infty, \infty) =\bigcup_{k=1}^{\infty}[-\infty,k)$$, $$(a, \infty] =\bigcup_{k=1}^{\infty} [a+\frac{1}{k},\infty]$$, $$(-\infty, \infty] =\bigcup_{k=1}^{\infty} (-k,\infty]$$, and the sets $$\{\infty\} =[-\infty, \infty] \setminus [-\infty, \infty)$$, $$\{-\infty\} =[-\infty, \infty] \setminus (-\infty, \infty]$$, $$\{-\infty,+\infty\}=\{-\infty\}\cup\{+\infty\}$$.

So, since every open subset of $$\mathbb{R}$$ can be written as a countable union of disjoint open intervals we have that $$\tau$$ is a $$\sigma$$-algebra on $$[-\infty,\infty]$$ that contains all the open subsets of $$\mathbb{R}$$.

Let $$\tau_{|\mathbb{R}}:=\{B\cap\mathbb{R}:B\in\tau\}$$: we claim that $$\tau_{|\mathbb{R}}$$ is a $$\sigma$$-algebra on $$\mathbb{R}$$.

• $$\emptyset\in\tau$$ and $$\emptyset\cap\mathbb{R}=\emptyset$$ so $$\emptyset\in\tau_{|\mathbb{R}}$$;

• $$E\in\tau_{|\mathbb{R}}\Rightarrow E=B\cap\mathbb{R}$$ for some $$B\in\tau$$ so $$\mathbb{R}\setminus E=\mathbb{R}\setminus (B\cap\mathbb{R})=(\mathbb{R}\setminus B) \cup (\mathbb{R}\setminus\mathbb{R})=\mathbb{R}\setminus B=(\mathbb{R}\setminus B)\cap \mathbb{R}$$ and $$\mathbb{R}, B\in\tau\overset{(2.25b)}{\Rightarrow}\mathbb{R}\setminus B\in\tau$$ so $$\mathbb{R}\setminus E\in\tau_{|\mathbb{R}}$$;

• $$E_1, E_2,\dots\in\tau_{|\mathbb{R}}\Rightarrow E_k=B_k\cap\mathbb{R},\ B_k\in\tau,\ k\geq 1$$. $$\bigcup_{k=1}^{\infty}E_k=\bigcup_{k=1}^{\infty}(B_k\cap\mathbb{R})=(\bigcup_{k=1}B_k)\cap\mathbb{R}$$ and $$\bigcup_{k=1}B_k\in\tau$$ so $$\bigcup_{k=1}^{\infty}E_k\in\tau_{|\mathbb{R}}$$.

So, $$\tau_{|\mathbb{R}}$$ is a $$\sigma$$-algebra on $$\mathbb{R}$$, as desired.

Now, if $$O$$ is an open subset of $$\mathbb{R}$$ then as we said before $$O\in\tau$$ and since $$O=O\cap\mathbb{R}$$ we have that $$O\in\tau_{|\mathbb{R}}$$ so $$\tau_{|\mathbb{R}}$$ is a $$\sigma$$-algebra on $$\mathbb{R}$$ containing all the open subsets of $$\mathbb{R}$$: since the set $$\mathcal{B}$$ of Borel subsets of $$\mathbb{R}$$ is by definition the smallest such $$\sigma$$-algebra we have that $$\mathcal{B}\subset\tau_{|\mathbb{R}}$$ and since $$\tau_{|\mathbb{R}}\subset\tau$$ ($$A\in\tau_{|\mathbb{R}}\Rightarrow A=B\cap\mathbb{R},\ B\in\tau$$ and since $$\mathbb{R}\in\tau$$ too we have that $$A=B\cap\mathbb{R}\in\tau$$) we have that $$\mathcal{B}\subset\tau$$.

Now, let $$C$$ be a Borel subset of $$[-\infty,\infty]$$: then by definition there exists a Borel subset $$B\subset\mathbb{R}$$ such that $$C=B$$ or $$C=B\cup\{-\infty\}$$ or $$C=B\cup\{\infty\}$$ or $$C=B\cup\{-\infty,\infty\}$$: $$B\in\tau$$ and also $$\{-\infty\},\{\infty\},\{-\infty,\infty\}\in\tau$$ so their possible unions are also all in $$\tau$$ thus $$C\in\tau$$ too so $$\tau$$ also contains all the Borel subsets of $$[-\infty,\infty]$$ so $$f^{-1}(B)\in\mathcal{S}$$ for all $$B\in [-\infty,\infty]$$ and in particular for all $$B\in\mathcal{B}$$ so $$f$$ is an $$\mathcal{S}$$-measurable function, as desired. $$\square$$

useful definitions

DEF. (Measurable function). Suppose $$(X,\mathcal{S})$$ is a measurable space. A function $$f:X\to [-\infty,\infty]$$ is called $$\mathcal{S}$$-measurable if $$f^{-1}(B)\in\mathcal{S}$$ for every Borel set $$B\subset[-\infty,\infty]$$

DEF. (Borel subset of $$[-\infty,\infty]$$. A subset of $$[-\infty,\infty]$$ is called a Borel set if its intersection with $$\mathbb{R}$$ is a Borel set. In other words a set $$C\subset [-\infty,\infty]$$ is a Borel set iff there exists a Borel set $$B\subset\mathbb{R}$$ such that $$C=B$$ or $$C=B\cup\{\infty\}$$ or $$C=B\cup \{-\infty\}$$ or $$C=B\cup\{\infty,\infty\}$$

• Looks good to me, except for this small detail: $\tau$ is a $\sigma$-algebra on $[-\infty,\infty]$, not $\mathbb R$. So when you say that $\tau$ must contain all Borel subsets of $\mathbb R$ because it contains all open subsets of $\mathbb R$ this is slightly incorrect reasoning, but the result is true. Commented Jul 26, 2021 at 14:50
• @Alphie thank you for your interest in my question; I have tried to amend the error (correction in boldtype in the text) and I would be grateful if you could check it out. Also, if you would then post your comment as an answer I would gladly accept it. Commented Jul 26, 2021 at 17:37

"Now, let $$B$$ be a Borel subset of $$\mathbb R$$: then $$B$$ is either an open subset of $$\mathbb R$$ or it can be obtained from open subsets of $$\mathbb R$$ by using the operations allowed in a σ-algebra (like complementation, difference, countable union and intersection)"
This statement is a bit loose, but can be formalized using transfinite induction. The main problem is that apparently there exist Borel sets that cannot be arrived at by a countable sequence of operations allowed in a $$\sigma$$-algebra. See here.
Define $$\tau|_\mathbb R:=\{B\cap \mathbb R:B\in \tau\}$$. Using the fact that $$\tau$$ is a $$\sigma$$-algebra on $$[-\infty,+\infty]$$, one checks that $$\tau|_\mathbb R$$ is a $$\sigma$$-algebra on $$\mathbb R$$. Moreover, $$\tau|_\mathbb R\subset \tau$$, since $$\mathbb R$$ is a Borel subset of $$[-\infty,+\infty]$$.
Now, you have argued that $$\tau$$ contains all open subsets of $$\mathbb R$$, but in fact these sets are all contained in $$\tau|_\mathbb R$$. Since the Borel $$\sigma$$-algebra on $$\mathbb R$$ is the smallest $$\sigma$$-algebra on $$\mathbb R$$ containing all open subsets of $$\mathbb R$$, and $$\tau|_\mathbb R$$ is such a $$\sigma$$-algebra, we deduce that $$\tau|_\mathbb R$$ contains all open subsets of $$\mathbb R$$. Finally, since $$\tau|_\mathbb R\subset \tau$$, it must be that $$\tau$$ contains all open subsets of $$\mathbb R$$ as well.