Is $f(x)=\begin{cases}\frac{1}{x}, x\neq 0\\ \infty, x=0 \end{cases}$ Borel-measurable? (Using Borel sets)

Question

Consider a function $f:\mathbb R\to\overline{\mathbb R}$ defined as $$f(x)=\begin{cases}\frac{1}{x}, x\neq 0\\ \infty, x=0 \end{cases}$$

Is $f$ Borel-measurable?

I followed the answer given here Prove $\phi(t)=1/t$ is Borel function. by martini

Let $a\in\mathbb R.$

For $a<0,$ \begin{align*} x \in f^{-1}([-\infty, a]) &\iff f(x) \le a \\ &\iff \frac 1x \le a\\ &\iff x \ge \frac 1a \&\ x < 0 \end{align*} Hence $f^{-1}([-\infty, a]) = [\frac 1a, 0)$, which is a Borel set.

For $a=0,$

\begin{align*} f^{-1}([-\infty,a])\\ &\iff f^{-1}([-\infty,a)\cup\{a\}\\ &\iff \frac 1x<a\cup\{a\}\\ \end{align*}

from here how to find the interval for $x$? like in the case $a<0$, $x$ was between certain values but in this case $\frac 1x<a\implies \frac 1a<x$ which can't be since $a=0$?

In the answer mentioned was $(-\infty, 0]$

Something similar happens in the case $a>0$.

Answer 1

For $a\in \mathbb{R}$.

You correctly proved that, for $a<0$, $$f^{-1}([-\infty, a]) = [\tfrac 1a, 0)$$

which is Borel.

For $a=0$, since: $ x \in f^{-1}([-\infty, 0]) \iff x<0$. (Details: Since $0$ is not in the image of $f$, we have: $ x \in f^{-1}([-\infty, 0]) \iff x \in f^{-1}([-\infty, 0))\iff \frac{1}{x}<0 \iff x<0$). So $$f^{-1}([-\infty, 0]) = (-\infty, 0)$$ which is Borel.

For $a>0$, we have that, for $x<0$, it is true that $\frac{1}{x}\le a$. On the other hand, we know that for $x>0$, $\frac{1}{x}\le a$ if and only if $x\ge \frac{1}{a}$. So we have: $$ f^{-1}([-\infty, a])= (-\infty, 0) \cup [\tfrac 1a, +\infty)$$ which is Borel.

Now, we need to inspect two other cases: $a=-\infty$ and $a = \infty$.

It is easy to see that

$$ f^{-1}([-\infty, -\infty])= \emptyset$$ which is Borel. And

$$ f^{-1}([-\infty, \infty])= \mathbb{R}$$ which is Borel.

So $f$ is Borel measurable.

Remark 1: Since the function $f$ is from $\mathbb R$ to $\overline{\mathbb R}$, to prove it is Borel measurable we need to prove that:

1. For any Borel set $B \subseteq \overline{\mathbb R}$, $f^{-1}(B)$ is a Borel set in $\mathbb R$.

It can be proved that this condition is equivalent to

2. For any $a \in \overline{\mathbb R}$, $f^{-1}([-\infty, a])$ is a Borel set in $\mathbb R$.

(Atention: $a \in \overline{\mathbb R}$, not just $a \in \mathbb R$)

Of course, we could also work with $a \in \mathbb R$ and explicitly check $f^{-1}(\{-\infty\} )$ and $f^{-1}(\{\infty\} )$ to ensure they are Borel set. Any way, to prove that $f$ is Borel measurable, we need to consider the extremes points of $\overline{\mathbb R}$.

Remark 2: Since $[-\infty, -\infty] = \{-\infty\}$, and since there is no $x \in \mathbb{R}$ such that $f(x)= -\infty$, we have: $$ f^{-1}([-\infty, -\infty])= \emptyset$$

Since $[-\infty, \infty]= \overline{\mathbb R}$ and, for all $x \in \mathbb{R}$, $f(x) \in \overline{\mathbb R}$, we have that

$$ \mathbb{R} \subseteq f^{-1}([-\infty, \infty])$$

But, since $\mathbb{R}$ is the domain of $f$, we also have

$$f^{-1}([-\infty, \infty]) \subseteq \mathbb{R}$$

So we conclude: $$f^{-1}([-\infty, \infty]) = \mathbb{R}$$

Answer 2

It suffices to prove that if $U$ is open in $\overline {\Bbb R}$ then $f^{-1}U$ is Borel in $\Bbb R.$

Let $g$ be the restriction of $f$ to the domain $D=\Bbb R$ \ $\{0\}.$ Now $D$ is open in $\Bbb R$ and $g:D\to \Bbb R$ is continuous.

If $U$ is open in $\overline {\Bbb R}$ then $U\cap \Bbb R$ is open in $\Bbb R$ and $$f^{-1}U=A\cup B$$ where $$A=f^{-1}(U\cap \Bbb R)=g^{-1}(U\cap \Bbb R)$$ and $$B= f^{-1}(U \setminus \Bbb R).$$ Now $A$ is open in $D$ by continuity of $g,$ and $D$ is open in $\Bbb R,$ so $A$ is open in $\Bbb R.$ So $A$ is Borel in $\Bbb R.$

And $B=\emptyset$ or $B=\{0\}$ so $B$ is Borel in $\Bbb R.$

So $f^{-1}U=A\cup B$ is Borel in $\Bbb R.$

Answer 3

Note that $f$ does not take on the values $0,-\infty.$ Note also that on $(-\infty,0)$, $f(x)=1/x$ is a bijection; same for $(0,\infty).$ Below we look at all cases of $a\in[-\infty,\infty].$

$a=-\infty:$ $f^{-1}([-\infty,a])=f^{-1}(\{-\infty\})=\emptyset,$ so we're fine.

$-\infty<a<0:$ Here $f^{-1}([-\infty,a])=f^{-1}((-\infty,a]),$ which is the interval $[1/a,0).$

$a=0:$ $f^{-1}([-\infty,0])=f^{-1}((-\infty,0)).$ As mentioned above, this is the interval $(-\infty,0).$

$0<a<\infty:$ In this case

$$f^{-1}([-\infty,a]) = f^{-1}((-\infty,0)\cup (0,a])$$ $$ = f^{-1}((-\infty,0))\cup f^{-1}((0,a])) = (-\infty,0)\cup [1/a,\infty).$$

$a=\infty.$ Here the inverse image is all of $\mathbb R.$