Reciprocals and Quotients of Measurable Functions

Where a Reciprocal Is Defined

For real-valued functions, division requires attention to the set where the denominator vanishes. If \(g\) is measurable, its zero set is measurable, so its complement is a measurable domain on which \(1/g\) is defined. A quotient can then be made into a function on the whole space by specifying a value on the zero set. The choice of that value is a convention; it is not a value supplied by division.

Throughout, let \((X,\mathcal{M},\mu)\) be any measure space; no assumption of finiteness, sigma-finiteness, or completeness is needed. All functions in this tutorial are finite real-valued functions, and measurability is with respect to \(\mathcal{M}\) and the Borel sigma-algebra \(\mathcal{B}_{\mathbb{R}}\). The conclusions are pointwise measurability statements, not assertions up to \(\mu\)-almost everywhere equality. We use the earlier level-set criterion: a real-valued function is measurable if and only if the preimage of every open half-line \((t,\infty)\), \(t\in\mathbb{R}\), is measurable. We also use that level and equality sets of measurable real-valued functions are measurable.

Since \(Z\in\mathcal{M}\), also \(D\in\mathcal{M}\). To discuss measurability of a function defined only on \(D\), we equip \(D\) with the trace sigma-algebra \(\mathcal{M}|_D=\{D\cap E:E\in\mathcal{M}\}\). A subset measurable in this trace sigma-algebra is also a member of \(\mathcal{M}\), because \(D\cap E\in\mathcal{M}\).

Measurability of the Reciprocal

Proof. We check the half-line criterion on \(D\). Fix \(t\in\mathbb{R}\). If \(t>0\), then \(1/g(x)>t\) is possible only when \(g(x)>0\). For positive \(g(x)\), multiplying by \(g(x)\) preserves the inequality and gives \(1>t g(x)\). Thus \(\{x\in D:1/g(x)>t\}=\{x:0<g(x)<1/t\}\). If \(t=0\), the inequality \(1/g(x)>0\) holds exactly when \(g(x)>0\), so the preimage is \(\{g>0\}\). If \(t<0\), every \(x\in D\) with \(g(x)>0\) qualifies. For \(g(x)<0\), multiplying \(1/g(x)>t\) by the negative number \(g(x)\) reverses the inequality and gives \(1<t g(x)\), or \(g(x)<1/t\). Hence \(\{x\in D:1/g(x)>t\}=\{g>0\}\cup\{g<1/t\}\). Each displayed set is measurable by the level-set criterion, and each is a subset of \(D\). The half-line criterion on the measurable space \((D,\mathcal{M}|_D)\) now proves that \(1/g\) is measurable there. \(\square\)

The sign split is essential. The reciprocal function is continuous on \(\mathbb{R}\setminus\{0\}\), but it is not a continuous real-valued function on all of \(\mathbb{R}\). The preimage calculation makes explicit both the restricted domain and the direction of the inequalities.

Quotients on the Whole Space

To make a quotient a function on all of \(X\), we will assign the value zero wherever its denominator vanishes. First extend the reciprocal by the same convention. The following formulas show directly that this extended function is measurable, including on the zero set.

Proof. Fix \(t\in\mathbb{R}\) and use the half-line criterion. If \(t>0\), points in \(Z\) have \(r=0\) and do not belong to \(\{r>t\}\); the reciprocal formula gives \(\{r>t\}=\{0<g<1/t\}\). If \(t=0\), the zero set again contributes no points, and \(\{r>0\}=\{g>0\}\). If \(t<0\), every point of \(Z\) qualifies because \(r=0>t\). Points where \(g>0\) also qualify. Among points where \(g<0\), the sign-reversal calculation in the preceding proof gives \(1/g>t\) exactly when \(g<1/t\). Therefore \(\{r>t\}=Z\cup\{g>0\}\cup\{g<1/t\}\). In each case the preimage is a measurable set, since \(Z\) and the indicated strict level sets are measurable. Thus \(r\) is measurable. \(\square\)

Proof. Define the measurable extension \(r\) of \(1/g\) by the preceding theorem. At every \(x\in X\), the stated convention gives \(q(x)=f(x)r(x)\): on \(D\) this is \(f(x)/g(x)\), and on \(Z\) it is \(f(x)\cdot0=0\). The earlier measurable sums and products result states that the product of two finite real-valued measurable functions on the same measurable space is measurable. Applying it to \(f\) and \(r\) proves that \(q\) is measurable. \(\square\)

The same argument allows any fixed real value \(c\) on \(Z\), rather than zero: define the reciprocal extension to equal \(c\) on \(Z\), check its half-line preimages by adding \(Z\) precisely when \(c>t\), and multiply by \(f\) only if the desired quotient convention is still zero. For a quotient assigned an arbitrary value on \(Z\), it is simplest to define it directly by measurable restrictions to \(D\) and \(Z\). The zero convention above is especially convenient because it is exactly the product \(f r\). In every case, the value on \(Z\) is a specified extension, not a claim that division by zero is defined.

A Common Error and Its Correction

A frequent mistake is to solve an inequality involving \(1/g\) without first restricting the sign of \(g\). For instance, the incorrect calculation \(\{x:1/x>2\}=\{x:x>1/2\}\) both loses the sign condition and fails to reverse the inequality correctly. On the domain \(x\ne0\), the inequality \(1/x>2\) requires \(x>0\); multiplying by this positive \(x\) then gives \(1>2x\), so the correct set is \((0,1/2)\). The point \(x=1/4\) verifies inclusion because \(1/(1/4)=4>2\), while \(x=1\) verifies exclusion because \(1/1=1\) is not greater than \(2\). When the multiplying quantity is negative, the inequality reverses, which is why reciprocal preimages must be split by sign.

These results establish measurability, not integrability or boundedness. They make no assertion that a quotient has a finite integral, and they do not remove the need to specify what happens where the denominator is zero. The practical sequence is: identify the measurable zero set, establish measurability on its complement, state the convention on the zero set, and then apply the measurable product rule when the convention is zero.