A Diagnostic Approach to Measurability
For an explicit formula, the question “is this function measurable?” is best answered by identifying its domain and computing inverse images of sets that determine measurability. A formula may be continuous on its actual domain but fail to define a function on the larger set one first had in mind. A sequence may have measurable terms and a measurable pointwise limit, while an uncountable supremum of measurable functions may fail to be measurable. These are distinct issues, and each has a direct diagnostic.
We use two settings. In the general criterion below, \((X,\mathcal M)\) is a measurable space and \(f:X\to\mathbb R\) is real-valued. The examples on the real line use either \((\mathbb R,\mathcal B_{\mathbb R})\) or \((\mathbb R,\mathcal L,m)\), where \(m\) is Lebesgue measure; these are sigma-finite measure spaces, and \((\mathbb R,\mathcal L,m)\) is complete. A function is measurable if the inverse image of every Borel subset of \(\mathbb R\) belongs to the sigma-algebra on its domain.
Proof. If \(f\) is measurable, then \((q,\infty)\) is open and hence Borel, so \(\{f>q\}=f^{-1}((q,\infty))\in\mathcal M\) for each rational \(q\). Conversely, suppose every such strict level set is measurable. For any real \(b\), density of the rationals gives \(\{f<b\}=\bigcup_{q\in\mathbb Q,\ q<b}\{f\le q\}\). Indeed, if \(f(x)<b\), choose a rational \(q\) with \(f(x)\le q<b\); and if \(f(x)\le q<b\), then \(f(x)<b\). Each \(\{f\le q\}\) is the complement of \(\{f>q\}\), so \(\{f<b\}\in\mathcal M\). Thus, for rational \(a<b\), the inverse image of \((a,b)\) is \(\{f>a\}\cap\{f<b\}\), which is measurable. Every open subset of \(\mathbb R\) is a countable union of open intervals with rational endpoints: take all such intervals contained in the open set. Its inverse image is therefore measurable. Since the Borel sigma-algebra is generated by the open sets, \(f\) is measurable. \(\square\)
The criterion gives a practical choice: compute \(\{f>q\}\), rather than trying to handle every Borel set separately. Its quantifier over all rational thresholds matters. Checking only a convenient finite or integer list is not enough.
Worked Example: A Rational Expression with an Exact Level-Set Test
On \((\mathbb R,\mathcal B_{\mathbb R})\), consider \(f(x)=\dfrac{x^2}{1+x^2}\). The denominator is positive for every real \(x\), so this formula defines a real-valued function on all of \(\mathbb R\). For a rational threshold \(q\), solve \(f(x)>q\). If \(q<0\), then \(f(x)\ge0>q\), so the level set is \(\mathbb R\). If \(q\ge1\), then \(x^2<1+x^2\), hence \(f(x)<1\le q\), and the level set is empty. For \(0\le q<1\), multiply by the positive number \(1+x^2\):
At \(q=0\), this says \(|x|>0\), so the level set is \(\mathbb R\setminus\{0\}\), as expected. For \(0<q<1\), it is the union of the two open rays outside the displayed endpoints. All these sets are Borel, so the criterion proves that \(f\) is Borel measurable. Check: at \(q=\tfrac12\), the formula gives \(|x|>1\). Substitution confirms \(x^2/(1+x^2)>\tfrac12\) exactly when \(2x^2>1+x^2\), or \(x^2>1\).
Check the Domain Before the Formula
A formula containing a logarithm, a denominator, or a root does not automatically define a function on the whole ambient space. Measurability is a property of a function on a specified measurable space, so the domain must be settled first. If the natural domain is \(D\subseteq\mathbb R\), its measurable structure can be taken to be the trace sigma-algebra \(\mathcal L|_D=\{E\cap D:E\in\mathcal L\}\) when working with Lebesgue measurability.
Worked Example: The Domain of a Logarithmic Formula
Let \(D=\mathbb R\setminus\{0\}\), with the trace of \(\mathcal L\), and define \(g:D\to\mathbb R\) by \(g(x)=\ln(x^2)\). The domain is necessary because \(\ln(0)\) is undefined. For any real \(a\),
The endpoints are excluded because the inequality is strict. This is a Borel subset of \(\mathbb R\), and hence belongs to the trace sigma-algebra on \(D\). The rational strict level-set criterion shows that \(g\) is measurable on \(D\). Check: when \(a=0\), the set is \((-\infty,-1)\cup(1,\infty)\); on it \(x^2>1\), so \(\ln(x^2)>0\), while at \(x=\pm1\) the value is \(0\) and the strict inequality fails.
A frequent domain error is to call \(x\mapsto\ln(x^2)\) a function on \(\mathbb R\) and then calculate its level sets there. The expression has no value at \(0\), so that claim does not define a function on \(\mathbb R\). One may instead use the domain \(D\), or explicitly assign a value at \(0\), for example by defining a new piecewise function. That is a different function and must be checked on its stated domain.
Sequences: Calculate the Limit, Then Use the Criterion
Earlier results on pointwise limits state that a pointwise limit of measurable real-valued functions is measurable when the limit exists as a real number at every point. For sequences, the hypotheses should be checked at the points where the limit is being asserted; the theorem is not a license to ignore undefined values or points where a finite limit fails to exist.
Worked Example: A Sequence with a Discontinuous Limit
On \((\mathbb R,\mathcal B_{\mathbb R})\), let \(f_n(x)=\dfrac{x^2}{x^2+1/n}=\dfrac{nx^2}{1+nx^2}\), for \(n\ge1\). The denominator \(1+nx^2\) is positive, and each \(f_n\) is continuous, hence Borel measurable. At \(x=0\), \(f_n(0)=0\) for every \(n\), so the limit is \(0\). If \(x\ne0\), then \(nx^2\to\infty\) and \(f_n(x)=1-\dfrac{1}{1+nx^2}\to1\). Therefore the pointwise limit is \(f(0)=0\) and \(f(x)=1\) for \(x\ne0\).
For a direct measurability check, if \(a<0\), then \(\{f>a\}=\mathbb R\); if \(a=0\), then \(\{f>a\}=\mathbb R\setminus\{0\}\); if \(0<a<1\), the same set \(\mathbb R\setminus\{0\}\) results; and if \(a\ge1\), the set is empty. Each is Borel, so \(f\) is measurable. Check: for \(a=\tfrac12\), \(f(x)>\tfrac12\) holds at every nonzero \(x\), since there \(f(x)=1\), and fails at \(0\), since \(f(0)=0\). This agrees with the calculated level set.
Pointwise convergence here does not mean uniform convergence on \(\mathbb R\). For example, at \(x=1/\sqrt n\), \(f_n(x)=1/2\), whereas the limit there is \(1\); thus the difference at that point is \(1/2\) for every \(n\). This does not affect the pointwise-limit measurability conclusion, which requires pointwise convergence, not uniform convergence.
Why Uncountable Operations Need Separate Attention
Proof. Fix \(x\in X\). If \(F(x)>a\), then not all values \(f_i(x)\) can be at most \(a\), since that would make their supremum at most \(a\). Hence \(f_i(x)>a\) for some \(i\), so \(x\in\bigcup_{i\in I}\{f_i>a\}\). Conversely, if \(f_i(x)>a\) for some \(i\), then \(F(x)\ge f_i(x)>a\). This proves the equality of sets. When \(I\) is countable, the right-hand side is a countable union of measurable sets. For an uncountable \(I\), sigma-algebra closure only guarantees closure under countable unions, so the equality alone does not guarantee measurability. \(\square\)
Worked Example: An Uncountable Supremum That Is Not Measurable
Work on \((\mathbb R,\mathcal L,m)\). Choose a non-Lebesgue-measurable set \(A\subseteq\mathbb R\), whose existence is a standard result, and for each \(t\in A\) define \(f_t=\chi_{\{t\}}\). Each singleton \(\{t\}\) is Borel, so each \(f_t\) is Lebesgue measurable. At a point \(x\in A\), the index \(t=x\) is present and \(f_x(x)=1\), while all these indicators take values at most \(1\); therefore \(\sup_{t\in A}f_t(x)=1\). At a point \(x\notin A\), every \(t\in A\) differs from \(x\), so every \(f_t(x)=0\), and the supremum is \(0\). Thus the supremum equals \(\chi_A\).
Its strict level set at threshold \(1/2\) is \(\{\sup_{t\in A}f_t>1/2\}=A\), which is not in \(\mathcal L\). Consequently the supremum is not Lebesgue measurable, even though every member of the family is. Check: at \(x\in A\) the supremum is exactly \(1>1/2\); at \(x\notin A\) it is exactly \(0\), so the strict level set is exactly \(A\), with no endpoint ambiguity.
Common Diagnostic Errors
Using too few thresholds. The rational-level criterion requires every rational threshold, not just integer thresholds. To see the failure concretely, on \((\mathbb R,\mathcal L,m)\) take a non-Lebesgue-measurable set \(A\) and define \(h=3/2\) on \(A\), \(h=7/4\) on \(A^c\). For every integer \(k\le1\), \(\{h>k\}=\mathbb R\); for every integer \(k\ge2\), \(\{h>k\}=\varnothing\). Those integer tests are all measurable. But \(\{h>13/8\}=A^c\), since \(3/2<13/8<7/4\). The complement of a nonmeasurable set is nonmeasurable, so \(h\) is not measurable. The correction is to test all rational thresholds, or to use another complete criterion, not to infer measurability from a sparse collection of level sets.
Taking an uncountable union as if it were countable. The level-set formula for a supremum is exact, but measurability follows automatically from it only when the union is countable. The singleton-indicator example shows that the gap is real, not merely a missing step in a proof.
Ignoring where a formula is defined. Before computing preimages, specify the domain and its sigma-algebra. An expression undefined at even one point is not yet a function on a domain containing that point. Once the domain is correct, preimages or an applicable continuity or limit result can settle the measurability question.
Check Your Understanding
For each question, identify the domain and the exact criterion you would use before concluding measurability.
- For \(u(x)=\dfrac{x}{1+|x|}\) on \(\mathbb R\), compute \(\{u>q\}\) for \(q=-1,0,1\), paying attention to strictness.
- On \(D=(0,\infty)\), determine \(\{x:\sqrt{x}>a\}\) for \(a<0\) and \(a\ge0\), and explain why the domain matters.
- For \(v_n(x)=\dfrac{x}{1+n|x|}\) on \(\mathbb R\), calculate the pointwise limit and state which pointwise-limit result applies.
- Why does the rational strict level-set criterion not permit checking only integer thresholds? Construct or describe a counterexample.
- For an uncountable family \((g_i)_{i\in I}\), write the strict level set of \(\sup_i g_i\). What closure property would be needed to conclude measurability from that expression?