Why Is the Function f(x) = 1/x Borel Measurable?

The reciprocal function is Borel measurable wherever it is defined. The key point is that $f(x)=1/x$ is continuous on its natural domain, the real line with zero removed. If you need a function defined on all of $\mathbb{R}$, you must assign a value at zero; every choice gives a Borel-measurable extension.

The domain matters

The expression $1/x$ has no value at $x=0$. So the function described by the formula is properly written as

$$f:D\to\mathbb{R},\qquad f(x)=\frac{1}{x},\qquad D=\mathbb{R}\setminus\{0\}.$$

A function $f$ from a topological space to $\mathbb{R}$ is Borel measurable if the preimage $f^{-1}(U)$ is a Borel set for every open set $U\subseteq\mathbb{R}$. For the domain $D$, “Borel” means Borel in the subspace topology on $D$.

Proof using continuity

The reciprocal function is continuous at every $x_0\in D$. For example, if $x$ is close enough to $x_0$ that $|x-x_0|<|x_0|/2$, then $|x|>|x_0|/2$, and

$$\left|\frac{1}{x}-\frac{1}{x_0}\right| =\frac{|x-x_0|}{|x||x_0|} <\frac{2|x-x_0|}{|x_0|^2}.$$

Given any $\varepsilon>0$, choosing $|x-x_0|$ sufficiently small makes the final expression less than $\varepsilon$. This proves continuity at $x_0$; since $x_0$ was arbitrary, $f$ is continuous on $D$.

Now let $U$ be any open subset of $\mathbb{R}$. Continuity means $f^{-1}(U)$ is open in $D$. Every open set is Borel, so $f^{-1}(U)$ is Borel in $D$. Thus $f$ is Borel measurable.

This also proves measurability for all Borel target sets, not just open ones. The collection of target sets whose preimages are Borel is a sigma-algebra, and it contains every open set. Since the Borel sigma-algebra is generated by the open sets, it contains every Borel set.

A concrete preimage

Take the open interval $U=(-2,3)$. Solving $-2<1/x<3$ separately for negative and positive $x$ gives

$$f^{-1}((-2,3))=(-\infty,-1/2)\cup(1/3,\infty).$$

This is open in $D$ (and also open in $\mathbb{R}$), as expected. Splitting by sign matters: the reciprocal reverses inequalities on each side of zero, and the two sides of the domain must be considered separately.

If you assign a value at zero

Suppose you define $g:\mathbb{R}\to\mathbb{R}$ by $g(x)=1/x$ for $x\ne0$ and choose any real number $c$ for $g(0)$. For an open set $U$,

$$g^{-1}(U)=f^{-1}(U)\cup\begin{cases}\{0\},&c\in U,\\\varnothing,&c\notin U.\end{cases}$$

The preimage from the punctured domain is Borel in $\mathbb{R}$, and the singleton $\{0\}$ is closed, hence Borel. Their union is Borel. Therefore every such extension is Borel measurable. The extension need not be continuous at zero; measurability is a weaker condition.

Common mistakes

• Treating $1/x$ as defined at zero. The formula alone defines a function on $\mathbb{R}\setminus\{0\}$, not on all of $\mathbb{R}$.
• Checking only one side of zero. The domain has two components, and the reciprocal behaves differently on the positive and negative sides.
• Confusing continuity with measurability. Continuity is a sufficient reason for Borel measurability, but it is not necessary. A function can be Borel measurable and discontinuous.
• Using Lebesgue measurability without specifying it. Borel measurability asks for Borel preimages; Lebesgue measurability uses the larger Lebesgue sigma-algebra. A Borel-measurable function is also Lebesgue measurable on the usual real line.