Riemann’s integral has a defect that no amount of care repairs. Enumerate the rationals of \([0,1]\) as \(q_1,q_2,\dots\) and let \(f_n\) be the indicator function of \(\{q_1,\dots,q_n\}\) . Each \(f_n\) is zero except at finitely many points, so it is Riemann integrable with integral \(0\) . The sequence increases, it is bounded by \(1\) , and it converges at every point. Its limit is the indicator of the rationals, which is not Riemann integrable: every subinterval contains points where it is \(1\) and points where it is \(0\) . The class of Riemann integrable functions is not closed under even the gentlest kind of limit.

Lebesgue’s remedy starts from a different way of slicing. To integrate \(f\) , do not cut the domain into intervals; cut the range into intervals, and ask how much of the domain is sent into each slice. That requires the sets \(\{x:a<f(x)\le b\}\) to have a measure. The functions for which they do are the measurable functions, and this chapter shows that they form a class closed under every countable operation of analysis, and that each of them is a limit of functions taking only finitely many values. On those two facts the integral of Chapter 4 is built.

3.1 Functions that pull measurable sets back

The definition has the same form as the definition of continuity by open sets, with “measurable” in place of “open”. As with continuity, it is the preimage that matters. For a real function \(f\) and a real number \(a\) we write \(\{f>a\}\) for \(\{x\in X:f(x)>a\}=f^{-1}((a,\infty))\) , and similarly \(\{f\le a\}\) , \(\{f=g\}\) , and so on.

Checking every Borel set would be hopeless. It is never necessary.

Three families of measurable functions come for free. A continuous function \(f:\mathbb R\to\mathbb R\) is Borel measurable, because \(\{f>a\}\) is open. A monotone function is Borel measurable, because \(\{f>a\}\) is an interval or a ray. And the indicator \(\mathbf 1_A\) is measurable exactly when \(A\in\mathcal A\) , because \(\{\mathbf 1_A>a\}\) is \(\emptyset\) , \(A\) , or \(X\) according as \(a\ge1\) , \(0\le a<1\) , or \(a<0\) .

It is convenient to allow the values \(\pm\infty\) . A function \(f:X\to[-\infty,\infty]\) is called measurable if \(\{f>a\}\in\mathcal A\) for all real \(a\) ; then \(\{f=\infty\}=\bigcap_n\{f>n\}\) and \(\{f=-\infty\}=\bigcap_n\{f\le-n\}\) are measurable too.

The order of the hypotheses matters. If \(\varphi\) is only Lebesgue measurable, then \(\varphi^{-1}(B)\) is a Lebesgue measurable set that may fail to be Borel, and \(f^{-1}\) of such a set need not be measurable; so \(\varphi\circ f\) can fail to be Lebesgue measurable even when \(f\) is continuous. The standard counterexample uses the Cantor function of Example 10.4 to carry a non-measurable set into the Cantor set by a homeomorphism. For that reason “measurable” for a function on \(\mathbb R\) with values in \(\mathbb R\) always means: Borel sets in the target.

3.2 Arithmetic and lattice operations

In the proof for the sum, the set \(\{f+g>a\}\) is naturally an uncountable union over all real thresholds, and the density of the rationals reduces it to a countable one. The same device will recur whenever a condition involving a real parameter has to be made countable.

3.3 Limits of sequences

This is the property that the Riemann theory lacks and that everything afterwards depends on.

The word “sequence” cannot be removed. Let \(V\) be the non-measurable set of Theorem 2.15 and for each \(t\in V\) let \(f_t=\mathbf 1_{\{t\}}\) . Each \(f_t\) is Borel measurable, yet \(\sup_{t\in V}f_t=\mathbf 1_V\) is not Lebesgue measurable. Countable operations preserve measurability; uncountable ones need a separate argument.

3.4 Simple functions and approximation from below

Simple functions are the finite linear combinations of indicators of measurable sets, and sums and products of simple functions are simple. They are to Lebesgue’s integral what step functions are to Riemann’s, with one decisive difference: the sets \(A_j\) need not be intervals. The indicator of the rationals is a simple function.

Two separate errors are being controlled. The quantization error is less than the mesh \(2^{-n}\) wherever the cut-off is inactive. The truncation error, the amount by which \(f\) exceeds \(n\) , is eliminated at each point once \(n\) passes \(f(x)\) , but not uniformly unless \(f\) is bounded. The mesh is dyadic so that each grid refines the previous one, and that is what makes the sequence increase.

Example 3.3: Dyadic approximants of a linear function

Let \(f(x)=3x\) on \([0,1]\) with Lebesgue measure. Since \(f\) is increasing, the level set \(\{k2^{-n}\le f<(k+1)2^{-n}\}\) is the interval \([\tfrac{k}{3\cdot2^n},\tfrac{k+1}{3\cdot2^n})\) , of length \(\tfrac1{3\cdot2^n}\) .

Level one. The mesh is \(\tfrac12\) and the cut-off is \(1\) . So \(\varphi_1=0\) on \([0,\tfrac16)\) , \(\varphi_1=\tfrac12\) on \([\tfrac16,\tfrac13)\) , and \(\varphi_1=1\) on \([\tfrac13,1]\) , where \(f\ge1\) . Weighting each value by the length of its level set,

\[ 0\cdot\tfrac16+\tfrac12\cdot\tfrac16+1\cdot\tfrac23=\tfrac34 . \]

Level two. The mesh is \(\tfrac14\) and the cut-off is \(2\) . Now \(\varphi_2=\tfrac k4\) on \([\tfrac k{12},\tfrac{k+1}{12})\) for \(k=0,\dots,7\) , and \(\varphi_2=2\) on \([\tfrac23,1]\) . The weighted sum is

\[ \frac1{12}\cdot\frac{0+1+\dots+7}{4}+2\cdot\frac13=\frac7{12}+\frac23=\frac54 . \]

Level three. The mesh is \(\tfrac18\) and the cut-off \(3\) is reached only at the single point \(x=1\) . So \(\varphi_3=\tfrac k8\) on \([\tfrac k{24},\tfrac{k+1}{24})\) for \(k=0,\dots,23\) , and the weighted sum is \(\tfrac1{24}\cdot\tfrac{0+1+\dots+23}{8}=\tfrac{23}{16}\) .

The three numbers \(\tfrac34\) , \(\tfrac54\) , \(\tfrac{23}{16}\) increase towards \(\tfrac32\) , the area under the graph. At levels one and two most of the shortfall comes from truncation; from level three on only the quantization error remains, and it is \(\tfrac1{16}\) , half the mesh. Figure 3.1 shows the three approximants. Chapter 4 will define the integral of \(f\) as the limit of such weighted sums.

Theorem 3.9 is the source of a method of proof used so often that it has no need of a name. To show that a statement holds for all nonnegative measurable functions: prove it for indicator functions; extend it to simple functions by linearity; extend it to all nonnegative measurable functions by taking increasing limits. Each stage requires that the statement be preserved under the corresponding operation, and Chapter 4 supplies the theorem that handles the third.

3.5 Almost uniform convergence

Pointwise convergence and uniform convergence look far apart. On a space of finite measure they are separated only by sets of arbitrarily small measure.

The hypothesis of finite measure enters at exactly one point, the appeal to continuity from above, and Example 1.3 has already shown what happens to that when the measure is infinite.

Exercises

1. Let \(X=\{1,2,3,4\}\) carry the \(\sigma\) -algebra with atoms \(\{1,2\}\) and \(\{3,4\}\) . Decide which of \(f(x)=x\) , \(g(x)=\lceil x/2\rceil\) , and \(h(x)=(x-\tfrac52)^2\) are measurable, giving a level set that fails when the answer is no.

   Answer

   Only \(g\) is measurable: it equals \(1\) on \(\{1,2\}\) and \(2\) on \(\{3,4\}\) , so it is constant on atoms. For \(f\) , the level set \(\{f>1\}=\{2,3,4\}\) is not a union of atoms. For \(h\) , the values at \(1,2,3,4\) are \(\tfrac94,\tfrac14,\tfrac14,\tfrac94\) , and \(\{h>1\}=\{1,4\}\) is not a union of atoms.

2. Prove directly from Proposition 3.2 that \(f:X\to\mathbb R\) is measurable if and only if \(\{f\le a\}\) is measurable for every rational \(a\) , naming the generating family you use.

   Hint

   The rays \((-\infty,a]\) with rational \(a\) generate \(\mathcal B(\mathbb R)\) by Proposition 1.7, and \(\{f\le a\}=f^{-1}((-\infty,a])\) . Proposition 3.2 gives one direction; the other is the definition of measurability applied to a Borel set.

3. (a) Prove that a monotone function \(f:\mathbb R\to\mathbb R\) is Borel measurable. (b) Prove that the derivative of an everywhere differentiable function \(F:\mathbb R\to\mathbb R\) is Borel measurable, although it need not be continuous.

   Solution

   (a) Suppose \(f\) is nondecreasing; the other case follows by considering \(-f\) . Fix \(a\) and let \(S=\{f>a\}\) . If \(x\in S\) and \(y>x\) , then \(f(y)\ge f(x)>a\) , so \(y\in S\) . A subset of \(\mathbb R\) that contains every point to the right of each of its points is \(\emptyset\) , \(\mathbb R\) , or a ray \((c,\infty)\) or \([c,\infty)\) with \(c=\inf S\) . Each of these is a Borel set, so \(f\) is Borel measurable by Corollary 3.3.

   (b) Since \(F\) is differentiable it is continuous. For each \(n\) the function

   \[ g_n(x)=n\bigl(F(x+\tfrac1n)-F(x)\bigr) \]

   is continuous, hence Borel measurable. By the definition of the derivative, \(g_n(x)\to F'(x)\) for every \(x\) , since \(g_n(x)\) is the difference quotient with increment \(\tfrac1n\) . A pointwise limit of a sequence of measurable functions is measurable by Theorem 3.6, so \(F'\) is Borel measurable. Continuity of \(F'\) was neither needed nor available: the function equal to \(x^2\sin(1/x)\) for \(x\ne0\) and to \(0\) at \(0\) is differentiable everywhere, with a derivative that is discontinuous at \(0\) .

4. Let \(f,g:X\to\mathbb R\) be measurable. Prove that \(\{f<g\}\) and \(\{f=g\}\) are measurable, and that the function equal to \(f\) on \(\{f<g\}\) and to \(g\) elsewhere is measurable.

   Solution

   For each \(x\) with \(f(x)<g(x)\) there is a rational \(q\) with \(f(x)<q<g(x)\) , and conversely. Hence

   \[ \{f<g\}=\bigcup_{q\in\mathbb Q}\bigl(\{f<q\}\cap\{g>q\}\bigr), \]

   a countable union of intersections of measurable sets. By symmetry \(\{g<f\}\) is measurable, and \(\{f=g\}\) is the complement of \(\{f<g\}\cup\{g<f\}\) . (Alternatively, \(g-f\) is measurable by Proposition 3.5 and the sets in question are \(\{g-f>0\}\) and \(\{g-f=0\}\) .)

   Let \(E=\{f<g\}\) and let \(h\) equal \(f\) on \(E\) and \(g\) on \(E^c\) . Then \(h=f\mathbf 1_E+g\mathbf 1_{E^c}\) . The indicators are measurable because \(E\) is, and products and sums of measurable functions are measurable by Proposition 3.5. In fact \(h=\min(f,g)\) , which gives a second proof, but the argument just given works for any measurable \(E\) and shows that measurable functions can be glued along measurable sets.

5. For \(f(x)=x^2\) on \([0,2]\) with Lebesgue measure, find the level sets of the dyadic approximants \(\varphi_1\) and \(\varphi_2\) of Theorem 3.9 and compute the two weighted sums of values times lengths, to four decimal places.

   Solution

   Level one. The mesh is \(\tfrac12\) and the cut-off is \(1\) . Since \(x^2\) is increasing on \([0,2]\) , the level sets are intervals: \(\varphi_1=0\) on \([0,\tfrac1{\sqrt2})\) , \(\varphi_1=\tfrac12\) on \([\tfrac1{\sqrt2},1)\) , and \(\varphi_1=1\) on \([1,2]\) , where \(x^2\ge1\) . The weighted sum is

   \[ \tfrac12\Bigl(1-\tfrac1{\sqrt2}\Bigr)+1\cdot1=\tfrac32-\tfrac1{2\sqrt2}\approx1.1464 . \]

   Level two. The mesh is \(\tfrac14\) and the cut-off is \(2\) . For \(k=0,\dots,7\) , \(\varphi_2=\tfrac k4\) on \([\tfrac{\sqrt k}2,\tfrac{\sqrt{k+1}}2)\) , an interval of length \(\tfrac12(\sqrt{k+1}-\sqrt k)\) , and \(\varphi_2=2\) on \([\sqrt2,2]\) . The weighted sum is

   \[ \sum_{k=0}^7\frac k4\cdot\frac{\sqrt{k+1}-\sqrt k}{2}+2(2-\sqrt2)\approx0.7902+1.1716=1.9617 . \]

   Both lie below \(\int_0^2x^2\,dx=\tfrac83\approx2.6667\) , and the shortfall is mostly truncation: at level two the function exceeds the cut-off on \([\sqrt2,2]\) , more than a quarter of the domain. The cut-off becomes inactive only at level four, since the maximum of \(f\) is \(4\) .

6. Let \(f:X\to\mathbb R\) be measurable. Prove that the function equal to \(1/f\) on \(\{f\ne0\}\) and to \(0\) on \(\{f=0\}\) is measurable, by computing its level sets.

   Hint

   Call the function \(g\) . For \(a>0\) , \(\{g>a\}=\{0<f<1/a\}\) . For \(a=0\) , \(\{g>0\}=\{f>0\}\) . For \(a<0\) , \(\{g>a\}=\{f\ge0\}\cup\{f<1/a\}\) . Each is built from level sets of \(f\) .

7. Give an example of a function \(f:\mathbb R\to\mathbb R\) that is not Lebesgue measurable although \(\lvert f\rvert\) and \(f^2\) are both constant, and say which part of Proposition 3.5 this shows cannot be reversed.

   Answer

   Let \(V\) be a non-measurable set and \(f=\mathbf 1_V-\mathbf 1_{V^c}\) . Then \(\lvert f\rvert=f^2=1\) everywhere, but \(\{f>0\}=V\) is not measurable. Measurability of \(f\) implies that of \(\lvert f\rvert\) and \(f^2\) , not conversely.

8. On \([0,1]\) let \(f_n(x)=nx/(1+nx)\) . Find the pointwise limit, show that convergence is not uniform on \((0,1]\) , and find the smallest \(n\) for which \(\lvert f_n-1\rvert<0.01\) on all of \([0.02,1]\) .

   Solution

   For \(x=0\) every \(f_n(0)=0\) . For \(x>0\) , \(f_n(x)=1-\frac1{1+nx}\to1\) . So the limit is \(f=\mathbf 1_{(0,1]}\) .

   On \((0,1]\) the error is \(\lvert f_n(x)-1\rvert=\frac1{1+nx}\) , which tends to \(1\) as \(x\to0\) for fixed \(n\) . Hence \(\sup_{(0,1]}\lvert f_n-1\rvert=1\) for every \(n\) and the convergence is not uniform there. (One can also see this without computing: each \(f_n\) is continuous on \([0,1]\) and the limit is not, so the convergence cannot be uniform on \([0,1]\) .)

   On \([\delta,1]\) the error \(\frac1{1+nx}\) is decreasing in \(x\) , so its maximum is \(\frac1{1+n\delta}\) . With \(\delta=0.02\) the requirement \(\frac1{1+0.02n}<0.01\) is \(1+0.02n>100\) , that is \(n>4950\) . The smallest such \(n\) is \(n=4951\) . This is Egorov’s theorem made explicit: after deleting the set \([0,0.02)\) of measure \(0.02\) , convergence is uniform, at a rate that deteriorates as the deleted set shrinks.

9. On \(\mathbb N\) with counting measure let \(f_n=\mathbf 1_{\{n,n+1,\dots\}}\) . Show that \(f_n\to0\) pointwise but not uniformly on any set whose complement has measure less than \(1\) , and explain which step of the proof of Theorem 3.11 fails.

   Hint

   A set whose complement has counting measure less than \(1\) has empty complement, so it is all of \(\mathbb N\) , and \(\sup_{\mathbb N}f_n=1\) . In the proof, with \(m=2\) the sets \(X\setminus E_{2,n}=\{n,n+1,\dots\}\) decrease to \(\emptyset\) but all have infinite measure, so continuity from above is unavailable; this is the tail computation of Example 1.3.

10. A student writes: “If \(f:\mathbb R\to\mathbb R\) is measurable then \(f^{-1}(B)\) is measurable for Borel \(B\) ; applying this to the function \(f^{-1}\) shows that \(f(A)\) is measurable for every Borel \(A\) .” Identify two distinct errors and give a counterexample on a finite space.

    Solution

    First error. The symbol \(f^{-1}(B)\) denotes the preimage of a set, which exists for every function. It does not presuppose an inverse function \(f^{-1}\) . A measurable function need not be injective or surjective, so “the function \(f^{-1}\) ” may not exist.

    Second error. Even when \(f\) is a bijection, measurability of \(f\) says nothing about measurability of its inverse. To apply the definition to the inverse map one would need to know that the inverse is measurable, which is exactly the statement that \(f\) sends measurable sets to measurable sets, the thing to be proved. The argument is circular.

    Counterexample. Take \(X=\{1,2,3\}\) with \(\mathcal A=\{\emptyset,\{1\},\{2,3\},X\}\) , as in Example 3.5, and let \(f\) be the identity from \((X,\mathcal P(X))\) to \((X,\mathcal A)\) . It is a measurable bijection, because every preimage is a subset of \(X\) and all subsets of the domain are measurable. Its inverse, the identity from \((X,\mathcal A)\) to \((X,\mathcal P(X))\) , is not measurable, and the image under \(f\) of the measurable set \(\{2\}\) is \(\{2\}\) , which is not in \(\mathcal A\) .

11. Let \(\{f_t\}_{t\in T}\) be an arbitrary family of continuous functions on \(\mathbb R\) , possibly uncountable, with \(\sup_tf_t(x)<\infty\) for every \(x\) . Prove that \(\sup_tf_t\) is Borel measurable, and explain why this does not contradict the example following Theorem 3.6.

    Solution

    Let \(g=\sup_{t\in T}f_t\) , which is real-valued by hypothesis. For any real \(a\) , \(g(x)>a\) holds exactly when \(f_t(x)>a\) for at least one \(t\) , so

    \[ \{g>a\}=\bigcup_{t\in T}\{f_t>a\}. \]

    Each \(\{f_t>a\}\) is open because \(f_t\) is continuous, and an arbitrary union of open sets is open. So \(\{g>a\}\) is open, hence a Borel set, for every \(a\) , and \(g\) is Borel measurable by Corollary 3.3. The union is uncountable, but openness, unlike measurability, survives uncountable unions.

    There is no conflict with the example after Theorem 3.6. There the functions were indicators of single points, whose level sets \(\{f_t>a\}\) are closed, not open; an uncountable union of closed sets can be any set whatever. The lesson is that an uncountable supremum of measurable functions is measurable only when some additional structure, here continuity, takes over the work that countability does in the theorem.

12. Prove that every measurable \(f:X\to[0,\infty]\) can be written as \(f=\sum_{k=1}^\infty c_k\mathbf 1_{E_k}\) with constants \(c_k>0\) and measurable sets \(E_k\) , not necessarily disjoint.

    Hint

    With \(\varphi_n\) as in Theorem 3.9 and \(\varphi_0=0\) , write \(f=\sum_{n\ge1}(\varphi_n-\varphi_{n-1})\) . Each difference is a nonnegative simple function, hence a finite sum of positive constants times indicators. List all the terms in a single sequence.

13. For measurable real functions \(f_n\) , show that the set of points where the sequence \(f_n(x)\) converges to a finite limit equals \(\bigcap_m\bigcup_n\bigcap_{j,k\ge n}\{\lvert f_j-f_k\rvert<\tfrac1m\}\) , and conclude that it is measurable.

    Hint

    A real sequence converges to a finite limit exactly when it is a Cauchy sequence (math.real-analysis.cauchy-completeness). Translate “for every \(m\) there is an \(n\) such that for all \(j,k\ge n\) ” into an intersection, a union, and an intersection, all countable.

14. Let \(f:\mathbb R\to\mathbb R\) be Lebesgue measurable. Use Theorem 2.10 and Corollary 3.10 to prove that there is a Borel measurable \(g\) with \(g=f\) almost everywhere with respect to \(\lambda\) .

    Hint

    First let \(f=\mathbf 1_E\) with \(E\in\mathcal L\) : by Theorem 2.10, \(E\) contains a Borel set \(H\) with \(\lambda(E\setminus H)=0\) , and \(g=\mathbf 1_H\) works. Extend to simple functions. For general \(f\) take simple \(\psi_n\to f\) from Corollary 3.10, Borel simple \(g_n=\psi_n\) off a null set \(N_n\) , and a Borel null set \(N\supseteq\bigcup_nN_n\) ; let \(g=\lim g_n\mathbf 1_{N^c}\) .

15. For \(f(x)=3x\) on \([0,1]\) as in Example 3.3, prove that for every \(n\ge3\) the weighted sum of \(\varphi_n\) equals \(\tfrac32-2^{-n-1}\) , and identify which of the two errors, truncation or quantization, this difference measures.

    Solution

    Let \(n\ge3\) . The cut-off \(n\) is at least the maximum value \(3\) of \(f\) , so \(\{f\ge n\}\) is empty for \(n>3\) and is the single point \(\{1\}\) , of measure zero, for \(n=3\) . Hence the weighted sum involves only the quantized levels. The level set \(\{k2^{-n}\le3x<(k+1)2^{-n}\}\) is an interval of length \(\frac1{3\cdot2^n}\) for \(k=0,1,\dots,3\cdot2^n-1\) , and these intervals fill \([0,1)\) . Writing \(M=3\cdot2^n\) , the weighted sum is

    \[ \sum_{k=0}^{M-1}\frac{k}{2^n}\cdot\frac1M=\frac{1}{2^nM}\cdot\frac{(M-1)M}{2}=\frac{M-1}{2^{n+1}}=\frac{3\cdot2^n-1}{2^{n+1}}=\frac32-2^{-n-1}. \]

    For \(n=3\) this is \(\tfrac32-\tfrac1{16}=\tfrac{23}{16}\) , in agreement with Example 3.3.

    The difference from \(\tfrac32\) is pure quantization error, since truncation is inactive. On each level set the function \(f-\varphi_n\) rises linearly from \(0\) to \(2^{-n}\) , so its average is half the mesh, \(2^{-n-1}\) , and the total length is \(1\) .

16. Let \(g:X\to\mathbb R\) be any function and let \(\mathcal A_g=\{g^{-1}(B):B\in\mathcal B(\mathbb R)\}\) , a \(\sigma\) -algebra by Exercise 1.2. Prove that if \(f:X\to\mathbb R\) is \(\mathcal A_g\) -measurable then \(f=h\circ g\) for some function \(h\) defined on \(g(X)\) .

    Solution

    Suppose \(x,y\in X\) satisfy \(g(x)=g(y)\) ; we show \(f(x)=f(y)\) . The singleton \(\{f(x)\}\) is a Borel set, so \(S=f^{-1}(\{f(x)\})\) belongs to \(\mathcal A_g\) by measurability of \(f\) . By the definition of \(\mathcal A_g\) there is a Borel set \(B\) with \(S=g^{-1}(B)\) . Since \(x\in S\) , we have \(g(x)\in B\) . Then \(g(y)=g(x)\in B\) , so \(y\in g^{-1}(B)=S\) , which says \(f(y)=f(x)\) .

    Now define \(h\) on \(g(X)\) by \(h(g(x))=f(x)\) . The previous paragraph shows that this does not depend on the choice of \(x\) with a given value \(g(x)\) , so \(h\) is a well-defined function, and \(f=h\circ g\) by construction. In words: a function measurable with respect to the \(\sigma\) -algebra generated by \(g\) can use only the information contained in the value of \(g\) . Example 3.1 is the finite case, where the atoms are the sets on which \(g\) is constant.

17. For subsets \(A_1,A_2,\dots\) of \(X\) show that \(\limsup_n\mathbf 1_{A_n}=\mathbf 1_{\limsup_nA_n}\) and \(\liminf_n\mathbf 1_{A_n}=\mathbf 1_{\liminf_nA_n}\) , and deduce the measurability of the two limit sets from Theorem 3.6.

    Answer

    \(\limsup_n\mathbf 1_{A_n}(x)=1\) exactly when \(\mathbf 1_{A_n}(x)=1\) for infinitely many \(n\) , that is, when \(x\in\limsup_nA_n\) ; otherwise it is \(0\) . Likewise \(\liminf_n\mathbf 1_{A_n}(x)=1\) exactly when \(x\in A_n\) for all large \(n\) . The indicators \(\mathbf 1_{A_n}\) are measurable, so their limsup and liminf are, and an indicator is measurable only if its set is.