Why Completeness Enters the Measure-Theoretic Argument
In the proof of the finite Radon–Nikodym theorem by von Neumann, one works with bounded linear functionals on \(L^2(\mu)\). The Hilbert-space representation theorem used afterward requires \(L^2(\mu)\) to be complete: every Cauchy sequence in its norm must converge to an element of the space. The important point here is not merely that a candidate limit can be written as a pointwise series. One must prove that the series defines a measurable function, that the function is square-integrable, and that convergence occurs in the \(L^2\) norm.
Throughout, \((X,\mathcal{M},\mu)\) is a finite measure space, so \(\mu(X)<\infty\). Functions may be real-valued or complex-valued. Two measurable functions are identified if they agree \(\mu\)-almost everywhere; thus elements of \(L^2(\mu)\) are equivalence classes, not individual pointwise formulas.
As usual, representatives can be chosen finite everywhere: if a representative is infinite on a measurable null set, set it equal to zero there. This measurable change does not change its equivalence class. We use the Cauchy–Schwarz inequality, Minkowski’s inequality for finite sums in \(L^2\), Monotone Convergence, and Fatou’s lemma. These standard integral results apply to measurable functions on any measure space; below, their hypotheses are specified where they are used.
The Completeness Theorem
Proof. Let \((f_n)\) be Cauchy in \(L^2(\mu)\). Choose measurable, finite-valued representatives. By the Cauchy condition, there are strictly increasing indices \(n_k\) such that, with \(g_k=f_{n_k}\),
For example, for each \(k\) choose an index \(N_k\) such that \(\|f_n-f_m\|_2\le2^{-k}\) whenever \(n,m\ge N_k\), and choose \(n_k\) recursively, strictly increasing, with \(n_k\ge N_k\). Then \(n_{k+1}\ge N_k\) as well, giving the displayed bound.
Put \(u_k=g_{k+1}-g_k\). Cauchy–Schwarz on the finite measure space gives \(\int |u_k|\,d\mu\le\mu(X)^{1/2}\|u_k\|_2\). This remains valid when \(\mu(X)=0\), with both sides zero. Hence
The functions \(|u_k|\) are nonnegative and measurable. By Monotone Convergence applied to their increasing partial sums, \(\int\sum_{k=1}^{\infty}|u_k|\,d\mu=\sum_{k=1}^{\infty}\int|u_k|\,d\mu<\infty\). A nonnegative measurable function with finite integral is finite almost everywhere: indeed, if it were infinite on a set of positive measure, its integral would be at least \(r\) times that measure for every positive integer \(r\). Thus the measurable set
has measure zero. On \(X\setminus N\), the series \(\sum_k u_k(x)\) converges absolutely, including in the complex case. Define \(h_m=\chi_{X\setminus N}\sum_{j=1}^{m}u_j\). Each \(h_m\) is measurable, and \(h_m\) converges pointwise everywhere to a measurable function \(h\): on \(X\setminus N\) it converges by absolute convergence, and on \(N\) it is identically zero. Set \(g=g_1+h\). This defines a measurable, finite-valued function on all of \(X\), with the series contribution set to zero on \(N\).
We next prove that \(g-g_1=h\) is in \(L^2(\mu)\), and estimate the tails. Fix \(k\ge1\). For \(m\ge k\), let \(T_{k,m}=\sum_{j=k}^{m}u_j\). Minkowski’s inequality, applied to this finite sum of \(L^2\) functions, gives \(\|T_{k,m}\|_2\le\sum_{j=k}^{m}\|u_j\|_2\le\sum_{j=k}^{m}2^{-j}\). On \(X\setminus N\), \(T_{k,m}\) tends to \(g-g_k\) as \(m\to\infty\), since the finite telescoping sum is \(g_{m+1}-g_k\) and the series defining \(g\) converges there. Fatou’s lemma applies to the nonnegative measurable functions \(|T_{k,m}|^2\); their pointwise limit is \(|g-g_k|^2\) almost everywhere. Therefore
In particular, \(g-g_1\in L^2(\mu)\); because \(g_1\in L^2(\mu)\), this implies \(g\in L^2(\mu)\). The same estimate shows \(\|g-g_k\|_2\le2^{1-k}\to0\). Finally, let \(\varepsilon>0\). Since \((f_n)\) is Cauchy, there is an \(N_0\) such that \(\|f_n-f_m\|_2<\varepsilon/2\) for \(n,m\ge N_0\). Choose \(k\) large enough that \(n_k\ge N_0\) and \(\|g-g_k\|_2<\varepsilon/2\). For \(n\ge N_0\), the triangle inequality gives \(\|f_n-g\|_2\le\|f_n-f_{n_k}\|_2+\|g_k-g\|_2<\varepsilon\). Thus \(f_n\to g\) in \(L^2(\mu)\), proving completeness. \(\square\)
The proof only needs finiteness of \(\mu(X)\) for the estimate from \(L^2\) to \(L^1\), which guarantees almost-everywhere absolute convergence of the difference series. Completeness of \(L^2\) itself holds for arbitrary measure spaces, but the proof above is the finite-measure argument suited to the theorem’s application.
Worked Examples: What the Proof Delivers
Worked Example: A Subsequence of Constant Functions
Take \(X=[0,1]\), \(\mathcal{M}\) the Lebesgue sigma-algebra, and \(\mu=m\). Let \(f_n(x)=1+1/n\). Since \(m([0,1])=1\), a constant \(c\) has \(L^2\) norm \(|c|\). Therefore \(\|f_n-f_m\|_2=|1/n-1/m|\), which tends to zero as \(n,m\to\infty\). The sequence is Cauchy.
Choose \(g_k=f_{2^k}\). Then \(u_k=g_{k+1}-g_k=2^{-(k+1)}-2^{-k}=-2^{-(k+1)}\), a constant function. Its norm is \(2^{-(k+1)}\), and \(\sum_{k=1}^{\infty}\|u_k\|_2=\sum_{k=1}^{\infty}2^{-(k+1)}=1/2\). The telescoping partial sums satisfy \(g_1+\sum_{j=1}^{m}u_j=1+2^{-(m+1)}\), so the limit is \(g(x)=1\). The tail from \(u_k\) onward is \(-2^{-k}\), giving \(\|g-g_k\|_2=2^{-k}\), exactly as the theorem’s tail bound predicts.
Check: Directly, \(\|f_n-g\|_2=|1/n|=1/n\to0\). The subsequence limit agrees with the direct pointwise and norm limit.
Worked Example: A Two-Point Space with Unequal Weights
Let \(X=\{a,b\}\), \(\mathcal{M}\) be all subsets of \(X\), and set \(\mu(\{a\})=2\), \(\mu(\{b\})=3\). This is a finite measure space with \(\mu(X)=5\). For \(f_n(a)=1+1/n\) and \(f_n(b)=-1+2/n\), the squared norm of a function with values \(s,t\) is \(2|s|^2+3|t|^2\). Thus, for positive integers \(n,m\),
As \(n,m\to\infty\), this tends to zero, so the sequence is Cauchy. Its limit is \(f(a)=1\), \(f(b)=-1\), and the squared norm of that limit is \(2(1)^2+3(-1)^2=5\). The difference has values \(1/n\) at \(a\) and \(2/n\) at \(b\), so \(\|f_n-f\|_2^2=2(1/n)^2+3(2/n)^2=14/n^2\), which tends to zero.
Check: Substituting the values into the weighted norm gives \(\|f_n-f\|_2=\sqrt{14}/n\). This confirms convergence in the actual measure-weighted norm, not the unweighted Euclidean norm.
Worked Example: Norm Convergence Need Not Be Pointwise Convergence of the Full Sequence
Again use \([0,1]\) with Lebesgue measure. For each integer \(k\ge1\) and \(0\le j<2^k\), let \(I_{k,j}=[j2^{-k},(j+1)2^{-k})\), and list the indicators \(\chi_{I_{k,j}}\) level by level, starting with \(k=1\). There are \(2^k\) intervals at level \(k\), each of measure \(2^{-k}\). The \(L^2\) norm of each corresponding indicator is \(\|\chi_{I_{k,j}}\|_2=(\int\chi_{I_{k,j}}\,dm)^{1/2}=2^{-k/2}\). As the list advances, its level \(k\) tends to infinity, so the norms of its terms tend to zero. Consequently the sequence converges to zero in \(L^2\).
However, every \(x\in[0,1)\) belongs to exactly one of the half-open intervals at each level \(k\). The listed function therefore takes value \(1\) at \(x\) infinitely often. At every level \(k\ge2\), it also takes value \(0\) at \(x\) for the other intervals at that level. Thus the sequence has no pointwise limit at any \(x\in[0,1)\); at \(x=1\), all listed indicators are zero. The convergence in \(L^2\) is exact, but it is not pointwise convergence of the full sequence.
Check: At level \(k=2\), each indicator has squared norm \(m(I_{2,j})=1/4\), hence norm \(1/2\), as the formula \(2^{-k/2}\) gives. At level \(k=4\), the norm is \(2^{-2}=1/4\). These norms decrease to zero while the pointwise values on \([0,1)\) keep alternating between \(0\) and \(1\).
A Measurability Pitfall and Its Correction
A common mistake is to establish convergence outside a null set and then say that the sum may be defined “arbitrarily” on that set. That is not valid on every measure space. If the sigma-algebra is not complete, a subset of a measurable null set need not be measurable. For instance, let \(\mathcal{M}\) be the Borel sigma-algebra on \([0,1]\), with Lebesgue measure restricted to Borel sets. The Cantor set is Borel and null, and it contains non-Borel subsets. The indicator of such a subset is not Borel measurable, so using it as an arbitrary extension can produce a function that is not an \(L^2\) function.
The correction in the proof is specific: the exceptional set \(N\) is measurable, and the series contribution is set equal to zero on \(N\). The approximating functions \(h_m=\chi_{X\setminus N}\sum_{j=1}^m u_j\) are measurable everywhere, and their pointwise limit is measurable everywhere. No completeness assumption on \((X,\mathcal{M},\mu)\) is needed. Also, the exceptional set is harmless for the norm estimates because it has measure zero, but that fact does not license an arbitrary, potentially nonmeasurable, definition there.
This completeness result is what permits \(L^2(\mu)\), with its inner product, to serve as a Hilbert space in the finite-measure argument. The subsequence method is useful beyond this application: it turns norm control of successive differences into an almost-everywhere-defined limit, then uses Fatou’s lemma to recover the required norm control.
Check Your Understanding
Use the proof’s hypotheses carefully, distinguishing pointwise statements from statements about equivalence classes.
- Why does \(\sum_k\|u_k\|_2<\infty\) imply \(\sum_k\int|u_k|\,d\mu<\infty\) in the finite-measure proof?
- Which named convergence theorem justifies \(\int\sum_k|u_k|\,d\mu=\sum_k\int|u_k|\,d\mu\), and what nonnegativity condition is needed?
- In the tail estimate, why is Fatou’s lemma applied to \(|T_{k,m}|^2\) rather than directly to \(T_{k,m}\)?
- Why does defining the series contribution as zero on the measurable exceptional set preserve measurability without assuming the measure space is complete?
- For the dyadic-indicator sequence, explain why convergence in \(L^2\) to zero does not imply pointwise convergence of the full sequence on \([0,1)\).