Products and Measurability

From Scalar Functions to Measurable Pairs

The previous tutorial showed that pointwise limits of measurable functions remain measurable. Products are another important way to build new functions, but measurability of each factor does not immediately answer the question: the product depends on the two values together. The useful step is to regard the pair of values as a function into the plane.

Throughout, \((X,\mathcal{F})\) is a measurable space, and real-valued functions are measurable with respect to the Borel sigma-algebra \(\mathcal{B}(\mathbb{R})\). The Borel sigma-algebra on \(\mathbb{R}^2\) is generated by open boxes with rational endpoints, as stated in the Rational-Box Generators Corollary. Each such box tests one condition on the first coordinate and one on the second. That makes it possible to check measurability of a pair using only measurability of its component functions.

The next theorem establishes that scalar measurability is enough to make this pair measurable. The rational endpoints matter: there are only countably many rational boxes, so their generated sigma-algebra can be tested using the defining countable closure properties of a sigma-algebra.

Proof. Let \(B=(p,q)\times(r,s)\) be an open box with rational endpoints \(p<q\) and \(r<s\). Its preimage is

Each set in this intersection is measurable: for example, \(\{p<f<q\}=\{f>p\}\cap\{f<q\}\), and threshold sets of a measurable real-valued function are measurable. Thus \(\Phi^{-1}(B)\in\mathcal{F}\) for every rational box \(B\).

Now define

The collection \(\mathcal{H}\) is a sigma-algebra on \(\mathbb{R}^2\): inverse images preserve complements and countable unions, and \(\mathbb{R}^2\in\mathcal{H}\) because its inverse image is \(X\). It contains every rational open box by the calculation above. Since those boxes generate \(\mathcal{B}(\mathbb{R}^2)\), every Borel set in \(\mathbb{R}^2\) belongs to \(\mathcal{H}\). Therefore \(\Phi\) is measurable. \(\square\)

The Product Theorem

Once the pair \(\Phi\) is known to be measurable, multiplication is a continuous operation on the plane. Define \(M:\mathbb{R}^2\to\mathbb{R}\) by \(M(u,v)=uv\). For any Borel set \(C\subseteq\mathbb{R}\), continuity implies \(M^{-1}(C)\) is Borel, by the Continuous Preimages of Borel Sets Theorem. The product function is precisely the composition \(M\circ\Phi\), so the Composition of Measurable Functions Theorem applies.

Proof. By the Measurability of a Pair Theorem, \(\Phi(x)=(f(x),g(x))\) is measurable as a map into \(\mathbb{R}^2\). The multiplication map \(M(u,v)=uv\) is continuous, hence measurable from \(\mathbb{R}^2\) with its Borel sigma-algebra to \(\mathbb{R}\) with its Borel sigma-algebra. The Composition of Measurable Functions Theorem now gives that \(M\circ\Phi\) is measurable. For every \(x\in X\),

Thus \(fg\) is measurable. \(\square\)

The same reasoning applies to other continuous operations. For instance, addition \((u,v)\mapsto u+v\) and subtraction \((u,v)\mapsto u-v\) are continuous, so sums and differences of measurable real-valued functions are measurable. The map \(u\mapsto cu\) is continuous for every fixed real \(c\), so constant multiples are measurable as well. These conclusions rely on both the measurability of the pair and the continuity of the operation being applied.

Products on Measurable Pieces

Products are also useful when a function is defined differently on different measurable regions. A common construction is to multiply a measurable function by an indicator: the product retains the function on a chosen set and sets it to zero elsewhere. This works whether or not the set is open, closed, or has positive measure; what matters here is measurability.

Why the Plane Argument Matters

It can be tempting to argue that \(f(x)g(x)\) is measurable simply because \(f\) and \(g\) are measurable. The conclusion is correct for real-valued functions, but the reason is not merely that the factors are individually measurable. The proof uses a particular feature of the real plane: its Borel sigma-algebra has a countable generating family of rational boxes. Their preimages can be checked using the component functions.

This countability also explains why it is important to establish measurability into \(\mathbb{R}^2\), rather than assuming that separate measurability automatically resolves every question about functions with several coordinates. In this setting, rational boxes provide the bridge. Once that bridge is established, continuous operations such as multiplication preserve measurability through composition.

A second common mistake is to infer more than the theorem says. Measurability of \(f\) and \(g\) guarantees measurability of \(fg\), but it does not guarantee continuity, boundedness, integrability, or any particular value for the measure of a level set. Each of those is a separate question requiring additional hypotheses or results. The theorem is a closure property: it ensures that multiplication does not leave the class of measurable real-valued functions.