From Pointwise Convergence to Measurable Level Sets
Measurability is preserved by many operations on functions, but taking a pointwise limit requires an argument: the limit need not be continuous, even when every function in the sequence is continuous. The useful observation is that a strict inequality for the limit can be detected by an eventual strict inequality for the sequence, provided we place a rational threshold strictly between the limit and the level being tested.
Let \((X,\mathcal{M})\) be any measurable space. A function \(g:X\to\mathbb{R}\) is measurable if \(g^{-1}(B)\in\mathcal{M}\) for every Borel set \(B\subseteq\mathbb{R}\). We use the half-line test: it is enough to check that \(\{x:g(x)>a\}\in\mathcal{M}\) for every \(a\in\mathbb{R}\). Indeed, open subsets of \(\mathbb{R}\) are countable unions of open intervals, and intervals can be formed from half-lines by countable set operations. Thus their preimages are measurable, which gives the Borel-preimage definition.
Suppose \(f_n:X\to\mathbb{R}\) are measurable and \(f_n(x)\to f(x)\) for every \(x\in X\), where \(f:X\to\mathbb{R}\) is real-valued everywhere. No measure is needed, and there are no finiteness or completeness assumptions: the argument uses only the sigma-algebra operations in \(\mathcal{M}\).
Exact Formulas for the Level Sets
Proof. Fix \(a\in\mathbb{R}\). We prove the displayed set identity in both directions. If \(x\in\{f>a\}\), then \(f(x)>a\). Since the rationals are dense in \(\mathbb{R}\), choose \(q\in\mathbb{Q}\) with \(a<q<f(x)\). Set \(\varepsilon=(f(x)-q)/2>0\). Pointwise convergence gives an \(N\) such that \(n\ge N\) implies \(|f_n(x)-f(x)|<\varepsilon\). Hence \(f_n(x)>f(x)-\varepsilon=(f(x)+q)/2>q\) for all \(n\ge N\). Therefore \(x\) belongs to the right-hand side.
Conversely, if \(x\) belongs to the right-hand side, there are \(q\in\mathbb{Q}\) with \(q>a\) and \(N\ge1\) such that \(f_n(x)>q\) for every \(n\ge N\). Taking the limit as \(n\to\infty\) gives \(f(x)\ge q\), so \(f(x)\ge q>a\). Thus \(x\in\{f>a\}\), proving the identity.
For each \(q\) and \(n\), the set \(\{f_n>q\}\) is in \(\mathcal{M}\) by the half-line test. The intersections over \(n\ge N\), and then the unions over \(N\) and the countable set \(\mathbb{Q}\cap(a,\infty)\), are measurable because \(\mathcal{M}\) is a sigma-algebra. Consequently \(\{f>a\}\in\mathcal{M}\) for every real \(a\). The half-line test now shows that \(f\) is measurable. If \(X\) is empty, the set identity and measurability conclusion hold as well, since every subset of \(X\) is empty. \(\square\)
The rational \(q\) is important. It gives a positive gap between \(a\) and the threshold that the sequence eventually exceeds. Without that gap, the limit could equal \(a\), even though every term of the sequence is strictly greater than \(a\).
Proof. If \(f(x)<a\), choose a rational \(q\) with \(f(x)<q<a\). With \(\varepsilon=(q-f(x))/2>0\), pointwise convergence implies that \(f_n(x)<f(x)+\varepsilon=(f(x)+q)/2<q\) for every sufficiently large \(n\). This places \(x\) in the right-hand side. In the other direction, eventual inequalities \(f_n(x)<q\) imply \(f(x)\le q<a\) on taking limits. The two inclusions prove the identity. Each set \(\{f_n<q\}\) is measurable by the lower half-line test, and the displayed operations are countable, so the formula also gives measurable lower level sets. \(\square\)
Worked Limits
Worked Example: Powers on the Unit Interval
On \(X=[0,1]\) with its Borel sigma-algebra, let \(f_n(x)=x^n\) for \(n\ge1\). Each \(f_n\) is continuous, hence measurable. If \(0\le x<1\), then \(x^n\to0\); at \(x=1\), \(x^n=1\) for every \(n\). Therefore $$ f(x)=\begin{cases}0,&0\le x<1,\\1,&x=1.\end{cases} $$
We compute the strict upper level sets, taking the inequality \(f(x)>a\) strictly. If \(a<0\), both possible values \(0\) and \(1\) exceed \(a\), so \(\{f>a\}=[0,1]\). If \(0\le a<1\), only the value \(1\) exceeds \(a\), so \(\{f>a\}=\{1\}\). If \(a\ge1\), neither value exceeds \(a\), so \(\{f>a\}=\varnothing\). These sets are Borel for every \(a\), verifying measurability directly. Check: at \(a=0\), the strict level set is \(\{1\}\), not all of \([0,1]\); at \(a=1\), it is empty.
Worked Example: Continuous Approximations to an Indicator
On \(\mathbb{R}\) with its Borel sigma-algebra, define $$ f_n(x)=\max\bigl(0,\min(1,nx)\bigr). $$
Each \(f_n\) is continuous, since it is obtained from the continuous function \(x\mapsto nx\) by taking maxima and minima with constants. If \(x\le0\), then \(f_n(x)=0\) for every \(n\). If \(x>0\), choose \(N\) with \(N\ge1/x\); for every \(n\ge N\), \(nx\ge1\), and \(f_n(x)=1\). Thus the pointwise limit is \(f=\chi_{(0,\infty)}\). For \(a<0\), \(\{f>a\}=\mathbb{R}\); for \(0\le a<1\), \(\{f>a\}=(0,\infty)\); for \(a\ge1\), \(\{f>a\}=\varnothing\). Each is Borel. Check: at \(x=-2\) and \(x=0\), every \(f_n\) is \(0\); at \(x=1/2\), \(f_2(x)=\max(0,\min(1,1))=1\), and all \(f_n(x)=1\) for \(n\ge2\). These values agree with the stated limit.
Worked Example: Seeing the Rational Threshold in the Formula
On \(X=\mathbb{R}\) with its Borel sigma-algebra, let \(f_n(x)=x+1/n\). Each function is continuous, and for each fixed \(x\), \(f_n(x)\to x\). We check the theorem's formula explicitly for a fixed \(a\in\mathbb{R}\).
Fix a rational \(q\). For any \(N\), the condition \(x\in\bigcap_{n=N}^{\infty}\{f_n>q\}\) means \(x+1/n>q\) for every \(n\ge N\). If \(x\ge q\), every such inequality holds because \(1/n>0\). If \(x<q\), choose \(n\ge N\) large enough that \(1/n\le q-x\); then \(x+1/n\le q\), so the condition fails. Therefore $$ \bigcap_{n=N}^{\infty}\{f_n>q\}=[q,\infty) $$ for every \(N\), and its union over \(N\) is still \([q,\infty)\). Taking the union over rational \(q>a\) gives \((a,\infty)\): every \(x>a\) has a rational \(q\) with \(a<q\le x\), while no \(x\le a\) belongs to any \([q,\infty)\) with \(q>a\). This is exactly \(\{x:f(x)>a\}\), since \(f(x)=x\). Check: for \(a=2\), the computed set is \((2,\infty)\); substituting the limit \(f(x)=x\) gives the same strict level set.
A Common Threshold Mistake
A tempting but incorrect formula is to test whether \(f_n>a\) eventually and infer \(f>a\). Take \(X=\{x_0\}\) with \(\mathcal{M}=\{\varnothing,X\}\), and define \(f_n(x_0)=1/n\). The sequence consists of measurable real-valued functions and converges everywhere to \(f(x_0)=0\). At \(a=0\), every term satisfies \(f_n(x_0)>0\), so the eventual condition holds, but \(f(x_0)>0\) is false. The conclusion fails because the terms approach the tested level itself.
The correction is to use \(q>a\). If the limit is strictly above \(a\), a rational \(q\) can be chosen strictly between them, and convergence then guarantees that all sufficiently late terms exceed \(q\). Conversely, if all sufficiently late terms exceed \(q>a\), their limit is at least \(q\), and is therefore strictly above \(a\). Both implications depend on the gap between the two thresholds.
The result concerns an everywhere pointwise limit with a real value at every point. The proof constructs the limit's level sets directly; it does not assert that pointwise convergence is uniform, nor does it use any measure or integration theorem. Its central reusable method is to translate a strict statement about a limit into an eventual strict inequality at a rational intermediate level.
Check Your Understanding
For each question, track whether the inequality is strict and whether the sequence condition is eventual.
- State the pointwise-limit measurability theorem, including the required range of the functions and where convergence must hold.
- Explain why a rational \(q\) with \(a<q<f(x)\) exists when \(f(x)>a\), and how pointwise convergence gives an eventual inequality above \(q\).
- For the sequence \(f_n(x)=x+1/n\), compute \(\{f>a\}\) directly and compare it with the rational-threshold formula.
- For \(f_n(x_0)=1/n\) on a one-point measurable space, identify exactly why eventual \(f_n(x_0)>0\) does not imply \(f(x_0)>0\).
- Does the theorem require \(\mu(X)<\infty\) or completeness of a measure space? Explain which properties the proof actually uses.