Why Continuity Preserves Measurability
A measurable real-valued function can have a complicated dependence on the point \(x\), but its preimages of Borel sets remain measurable. A continuous function on \(\mathbb{R}\) behaves well with respect to those preimages: the preimage of a Borel set under a continuous map is again Borel. These two facts fit together to show that applying a continuous real function to a measurable real-valued map preserves measurability.
Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space; \(\mu\) need not be finite, sigma-finite, or complete. A real-valued function \(g:X\to\mathbb{R}\) is measurable if \(g^{-1}(B)\in\mathcal{M}\) for every \(B\in\mathcal{B}_{\mathbb{R}}\), the Borel sigma-algebra on \(\mathbb{R}\). This definition concerns sets and does not require an integral to exist. We use the open-set criterion for measurability: it is enough to check preimages of open subsets of \(\mathbb{R}\).
Proof. We must show that \((\phi\circ g)^{-1}(B)\in\mathcal{M}\) for every \(B\in\mathcal{B}_{\mathbb{R}}\). First, \(\phi^{-1}(B)\) is Borel for every Borel set \(B\). To verify this, let \(\mathcal{C}=\{A\subseteq\mathbb{R}:\phi^{-1}(A)\in\mathcal{B}_{\mathbb{R}}\}\). Preimages preserve complements and countable unions, so \(\mathcal{C}\) is a sigma-algebra. Continuity of \(\phi\) says that \(\phi^{-1}(U)\) is open, hence Borel, for every open \(U\subseteq\mathbb{R}\). Thus \(\mathcal{C}\) contains all open sets and consequently contains the sigma-algebra they generate, which is \(\mathcal{B}_{\mathbb{R}}\). Now, for each \(B\in\mathcal{B}_{\mathbb{R}}\), the preimage identity gives \((\phi\circ g)^{-1}(B)=g^{-1}(\phi^{-1}(B))\). Since \(\phi^{-1}(B)\) is Borel and \(g\) is measurable, the right-hand side belongs to \(\mathcal{M}\). This proves the claim. \(\square\)
The proof uses no measure estimates: neither the values of \(\mu\) nor the size of \(X\) matter. The key point is the order of the two preimage operations. One first takes the preimage under the outer map \(\phi\), then the preimage under the inner map \(g\).
Worked Examples
Worked Example: A Polynomial of the Identity Map
On \((\mathbb{R},\mathcal{L},m)\), let \(g(x)=x\) and \(\phi(t)=t^2\). Show that \(x\mapsto x^2\) is measurable and compute the preimage of \((-\infty,4)\).
The identity map \(g(x)=x\) is measurable because \(g^{-1}(B)=B\in\mathcal{L}\) for every Borel set \(B\). The polynomial \(\phi(t)=t^2\) is continuous on \(\mathbb{R}\). The theorem therefore shows that \(\phi\circ g\), which is \(x\mapsto x^2\), is measurable. For the requested open set,
Indeed, \(x^2<4\) is equivalent to \(|x|<2\), which is equivalent to \(-2<x<2\). Check: at \(x=-2\) and \(x=2\), the value is \(4\), which is not in \((-\infty,4)\), so both endpoints are excluded. The computed preimage is an open interval and hence belongs to \(\mathcal{L}\), as required.
Worked Example: The Exponential of the Identity Map
On \((\mathbb{R},\mathcal{L},m)\), let \(g(x)=x\) and \(\phi(t)=e^t\). Show that \(x\mapsto e^x\) is measurable and determine the preimage of \((e^2,\infty)\).
The identity is measurable, and the exponential function is continuous on \(\mathbb{R}\). Thus the composition theorem gives measurability of \(x\mapsto e^x\). Since the exponential is strictly increasing, \(e^x>e^2\) holds exactly when \(x>2\). Therefore
Check: for \(x=2\), \(e^x=e^2\), which is not in the strict target interval; for \(x=3\), \(e^x=e^3>e^2\), and for \(x=1\), \(e^x=e<e^2\). The computed preimage is a Borel set, consistent with measurability.
Worked Example: Absolute Value of a Two-Valued Measurable Map
Let \(E\in\mathcal{M}\) and define \(g:X\to\mathbb{R}\) by \(g(x)=-2\) on \(E\) and \(g(x)=3\) on \(E^c\). Show that \(g\) and \(|g|\) are measurable, and compute the preimage of \((-\infty,5/2)\) under \(|g|\).
For any Borel set \(B\subseteq\mathbb{R}\), the preimage \(g^{-1}(B)\) is one of \(\varnothing\), \(E\), \(E^c\), or \(X\), according to whether \(B\) contains neither, just \(-2\), just \(3\), or both of the two values. Each possibility belongs to \(\mathcal{M}\), so \(g\) is measurable. The outer function \(\phi(t)=|t|\) is continuous, and hence \(|g|=\phi\circ g\) is measurable by the theorem. Its values are \(|-2|=2\) on \(E\) and \(|3|=3\) on \(E^c\). Since \(2<5/2\) and \(3\not<5/2\),
Check: the preimage is measurable because \(E\in\mathcal{M}\). The calculation also covers the degenerate cases \(E=\varnothing\) and \(E=X\): the preimage is respectively \(\varnothing\) and \(X\).
Preimages and a Common Mistake
The identity used in the theorem is worth keeping explicit. If \(B\subseteq\mathbb{R}\), then a point \(x\) belongs to \((\phi\circ g)^{-1}(B)\) exactly when \(\phi(g(x))\in B\). This is exactly the condition \(g(x)\in\phi^{-1}(B)\), so
The set \(B\) itself need not be the set to which we apply \(g^{-1}\); it is first pulled back through \(\phi\). For an open \(B\), continuity ensures that \(\phi^{-1}(B)\) is open, and the open-set criterion then gives the result directly.
Common mistake. For \(g(x)=x\), \(\phi(t)=t^2\), and \(B=(-\infty,4)\), it is incorrect to write \((\phi\circ g)^{-1}(B)=g^{-1}(B)=(-\infty,4)\). That calculation omits the preimage under \(\phi\). The correct calculation is \(\phi^{-1}(B)=(-2,2)\), followed by \(g^{-1}((-2,2))=(-2,2)\). The endpoint check matters: \(x=\pm2\) gives \(x^2=4\), and \(4\notin(-\infty,4)\).
Polynomials, the exponential, and absolute value are examples of continuous outer functions, so each can be applied to any measurable real-valued \(g\) by this theorem. The theorem does not claim that every function of a measurable map is measurable; its stated hypothesis is that the outer function is continuous. Nor does it require \(g\) to be bounded, integrable, or defined on a finite-measure space.
Continuous Composition and Almost-Everywhere Equality
Proof. The continuous composition theorem applies separately to \(g\) and \(h\), so both compositions are measurable. By the assumption \(g=h\) \(\mu\)-a.e., there is a set \(N\in\mathcal{M}\) with \(\mu(N)=0\) such that \(g(x)=h(x)\) for every \(x\in X\setminus N\). For each such \(x\), applying the same function \(\phi\) to both sides gives \(\phi(g(x))=\phi(h(x))\). Thus the compositions agree outside the null set \(N\), which is precisely equality \(\mu\)-a.e. No completeness assumption is needed, because the exceptional set is already taken to be a measurable null set. \(\square\)
This proposition explains why continuous transformations respect changes to a measurable function on a null set: outside the same exceptional set, the input values agree and therefore their transformed values agree. It is a statement about measurability and equality almost everywhere, not about integrability or convergence.
Check Your Understanding
In each question, distinguish the preimage of the target set under the outer function from the subsequent preimage under the measurable map.
- On \((\mathbb{R},\mathcal{L},m)\), let \(g(x)=x\) and \(\phi(t)=t^2\). Compute \((\phi\circ g)^{-1}((-\infty,9))\), including whether the endpoints are included.
- For an arbitrary measure space \((X,\mathcal{M},\mu)\), measurable \(g:X\to\mathbb{R}\), and continuous \(\phi:\mathbb{R}\to\mathbb{R}\), write the preimage identity for a Borel set \(B\) and explain why it gives a measurable set.
- Let \(E\in\mathcal{M}\), and let \(g=-1\) on \(E\) and \(g=4\) on \(E^c\). What is the preimage of \((-\infty,2)\) under \(|g|\)?
- Why does the continuous composition theorem require no assumption that \(\mu(X)\) is finite?
- If measurable \(g\) and \(h\) agree \(\mu\)-a.e., why must \(\phi\circ g\) and \(\phi\circ h\) agree \(\mu\)-a.e. for continuous \(\phi\)?