From Minkowski’s Inequality to Completeness
Minkowski’s inequality shows that the \(L^p\) norm satisfies the triangle inequality. It therefore defines a metric on \(L^p(X)\), the space of measurable functions with finite \(L^p\) norm, where functions equal almost everywhere are identified. The next question is whether this metric has the completeness property: must every Cauchy sequence of \(L^p\) functions converge to an element of \(L^p(X)\)?
The answer is yes for every \(1\leq p\leq\infty\), on an arbitrary measure space. The main construction is to select a subsequence whose successive differences have summable norms. For finite \(p\), this produces a pointwise convergent series outside a null set. For \(p=\infty\), the same summability gives a uniform bound on the tails outside a null set. In both cases, convergence of the subsequence is then extended to the original Cauchy sequence.
The norm and metric are defined on equivalence classes, not on particular representatives. Thus a statement about values of functions at individual points must be made using measurable representatives and must allow exceptional null sets. Countability is important: the union of countably many null sets is still null, so representatives can be adjusted on one exceptional set to make all the estimates needed in the proof hold at once.
The Summable-Increment Method
Let \((f_n)\) be Cauchy in \(L^p(X)\). For each positive integer \(j\), the Cauchy property allows us to choose an index \(n_j\), with the indices strictly increasing, so that consecutive chosen terms satisfy \(\|f_{n_{j+1}}-f_{n_j}\|_p\leq 2^{-j}\). The sequence of bounds has finite sum. This does not by itself show that the functions converge, but it gives control over all finite sums of their absolute successive differences.
The lemma supplies both parts needed for the finite-\(p\) proof: the series of absolute differences converges pointwise almost everywhere, and its sum is itself an \(L^p\) function. The pointwise convergence will define a candidate limit; the norm estimate will control the distance to that limit.
Completeness of \(L^p\) for Finite \(p\)
First suppose \(1\leq p<\infty\), and let \((f_n)\) be Cauchy in \(L^p(X)\). Choose a subsequence \((f_{n_j})\) such that \(\|f_{n_{j+1}}-f_{n_j}\|_p\leq2^{-j}\) for every \(j\). Choose measurable representatives \(g_j\) of the equivalence classes \(f_{n_j}\), and put \(u_j=g_{j+1}-g_j\). The summable-increment lemma applies to these differences. Thus \(G(x)=\sum_{j=1}^{\infty}|g_{j+1}(x)-g_j(x)|\) is finite almost everywhere and belongs to \(L^p(X)\).
Let \(E\) be the measurable set on which \(G\) is finite. For every \(x\in E\) and integers \(r>s\), the triangle inequality for real numbers gives \(|g_r(x)-g_s(x)|\leq\sum_{j=s}^{r-1}|g_{j+1}(x)-g_j(x)|\). The tails of a convergent series of nonnegative numbers tend to zero, so \((g_j(x))\) is a Cauchy sequence of real numbers for each \(x\in E\). Define \(g(x)=\lim_{j\to\infty}g_j(x)\) on \(E\), and set \(g(x)=0\) on \(X\setminus E\). Since \(E\) has full measure and \(g\) is a pointwise limit of measurable functions on \(E\), this defines a measurable function.
On \(E\), we have \(|g|\leq |g_1|+G\). Since \(g_1\) and \(G\) are in \(L^p(X)\), Minkowski’s inequality shows that \(|g_1|+G\) is in \(L^p(X)\). The pointwise bound therefore proves that \(g\in L^p(X)\). To estimate the error, for every \(m\) and almost every \(x\), \(|g(x)-g_m(x)|\leq H_m(x)\), where \(H_m=\sum_{j=m}^{\infty}|g_{j+1}-g_j|\). Applying Minkowski’s inequality to finite partial sums of this tail and then the Monotone Convergence Theorem to their \(p\)-th powers gives \(\|H_m\|_p\leq\sum_{j=m}^{\infty}2^{-j}=2^{1-m}\). By monotonicity of the integral, \(\|g-g_m\|_p\leq\|H_m\|_p\leq2^{1-m}\). Hence the chosen subsequence converges to \(g\) in \(L^p\).
It remains to pass from the subsequence to the original sequence. Let \(\varepsilon>0\). Since \((f_n)\) is Cauchy, there is \(N\) such that \(\|f_n-f_\ell\|_p<\varepsilon/2\) whenever \(n,\ell\geq N\). Choose \(j\) large enough that \(n_j\geq N\) and \(\|f_{n_j}-g\|_p<\varepsilon/2\). For any \(n\geq N\), Minkowski’s inequality now gives \(\|f_n-g\|_p\leq\|f_n-f_{n_j}\|_p+\|f_{n_j}-g\|_p<\varepsilon\). Thus \(f_n\to g\) in \(L^p(X)\), completing the finite-\(p\) proof.
The Essential-Supremum Case
For \(p=\infty\), the subsequence is chosen in the same way, with \(\|f_{n_{j+1}}-f_{n_j}\|_\infty\leq2^{-j}\). The key difference is that an \(L^\infty\) norm bound gives an almost-everywhere pointwise bound on each increment, rather than an integral bound on a sum of increments. For each \(j\), choose representatives \(g_j\) as before. The definition of essential supremum gives \(|g_{j+1}-g_j|\leq2^{-j}\) outside a null set \(N_j\). Because there are countably many increments, their exceptional sets have a null union \(N=\bigcup_{j=1}^{\infty}N_j\).
For every \(x\notin N\), the sequence \(g_j(x)\) is Cauchy: if \(r>s\), then \(|g_r(x)-g_s(x)|\leq\sum_{j=s}^{r-1}2^{-j}\). Define \(g(x)=\lim_{j\to\infty}g_j(x)\) outside \(N\), and set \(g=0\) on \(N\). This function is measurable. Also, outside \(N\), \(|g-g_1|\leq\sum_{j=1}^{\infty}2^{-j}=1\). Since \(g_1\) is essentially bounded, this proves \(g\in L^\infty(X)\). More generally, for every \(m\), \(|g-g_m|\leq\sum_{j=m}^{\infty}2^{-j}=2^{1-m}\) outside \(N\), so \(\|g-g_m\|_\infty\leq2^{1-m}\). The subsequence converges in \(L^\infty\). The Cauchy-sequence argument using Minkowski’s inequality from the finite-\(p\) proof applies unchanged with \(\|\cdot\|_\infty\), and proves convergence of the entire original sequence. This completes the proof. \(\square\)
Worked Examples
Worked Example: Indicators of Shrinking Intervals
On \([0,1]\) with Lebesgue measure, let \(f_n=\mathbf{1}_{[0,1/n]}\). If \(m>n\), then \([0,1/m]\subseteq[0,1/n]\), and the difference \(f_n-f_m\) is one on the interval between these endpoints and zero elsewhere, apart from endpoint choices that do not affect its integral. Therefore, for \(1\leq p<\infty\), \(\|f_n-f_m\|_p=(1/n-1/m)^{1/p}\leq n^{-1/p}\). This proves that \((f_n)\) is Cauchy in every finite-\(p\) space. Its limit in \(L^p\) is zero, since \(\|f_n\|_p=n^{-1/p}\to0\).
The same sequence is not Cauchy in \(L^\infty([0,1])\). For \(m>n\), the difference has absolute value one on a set of positive measure, so \(\|f_n-f_m\|_\infty=1\). This illustrates that a sequence can be Cauchy in every finite-\(p\) norm here but fail to be Cauchy in the essential-supremum norm.
Worked Example: A Series of Successive Increments
On \(\mathbb R\) with Lebesgue measure, let \(I_k=[k,k+1)\) and define \(f_n=\sum_{k=1}^{n}3^{-k}\mathbf{1}_{I_k}\). The intervals are disjoint and each has measure one. For \(m>n\) and \(1\leq p<\infty\), direct integration gives \(\|f_m-f_n\|_p^p=\sum_{k=n+1}^{m}3^{-kp}\leq\sum_{k=n+1}^{\infty}3^{-kp}=\frac{3^{-(n+1)p}}{1-3^{-p}}\). The right-hand side tends to zero as \(n\) increases, so the sequence is Cauchy. Its limit is \(f=\sum_{k=1}^{\infty}3^{-k}\mathbf{1}_{I_k}\), and the same calculation gives \(\|f-f_n\|_p=\left(\frac{3^{-(n+1)p}}{1-3^{-p}}\right)^{1/p}\).
For \(p=\infty\), the largest coefficient in the tail is \(3^{-(n+1)}\), and it occurs on \(I_{n+1}\), which has positive measure. Hence \(\|f-f_n\|_\infty=3^{-(n+1)}\). This sequence converges in all the stated norms, with the tail estimates displaying the summable-increment mechanism explicitly.
Worked Example: Polynomial Partial Sums in \(L^\infty\)
On \([0,1]\), define \(h_n(x)=\sum_{k=1}^{n}(x/2)^k\). The geometric-series formula gives the pointwise limit \(h(x)=x/(2-x)\). For every \(x\in[0,1]\), \(|h(x)-h_n(x)|=\frac{(x/2)^{n+1}}{1-x/2}\leq\frac{(1/2)^{n+1}}{1-1/2}=2^{-n}\). Thus \(\|h-h_n\|_\infty\leq2^{-n}\), so the partial sums converge in \(L^\infty([0,1])\). Since \([0,1]\) has measure one, the same pointwise bound also gives \(\|h-h_n\|_p\leq2^{-n}\) for every finite \(p\). This example verifies convergence directly; the completeness theorem guarantees that such limits exist even when they cannot be written down by a geometric-series formula.
Why the Null-Set and Subsequence Steps Matter
Completeness is not a claim that every Cauchy sequence of representatives converges at every point. It is a claim about convergence in norm of equivalence classes. In the finite-\(p\) argument, the series of absolute increments may fail to be finite on a null set; in the \(L^\infty\) argument, different norm estimates may initially each hold outside different null sets. The proofs handle these issues by defining the limit only on a common full-measure set and choosing any convenient values on the exceptional set.
A second common gap is to prove only that a subsequence converges and then stop. The Cauchy property is what transfers convergence from that subsequence to the full sequence: an arbitrary sufficiently late term is close to a sufficiently late subsequence term, which is itself close to the limit. The triangle inequality makes these two estimates combine. Both the subsequence construction and this final transfer are essential parts of the completeness proof.
From the Cauchy sequence, choose a subsequence with consecutive distances bounded by \(2^{-j}\).
For finite \(p\), use the summable-increment lemma; for \(p=\infty\), combine the countably many almost-everywhere increment bounds.
Bound the error by the tail of the summable increments.
Apply the Cauchy property and the triangle inequality to compare each late term with a late subsequence term and then with the limit.
Check Your Understanding
Use the proof to answer these questions.
- Why does the summable-increment lemma imply that the subsequence is pointwise Cauchy almost everywhere for finite \(p\)?
- Where is the Monotone Convergence Theorem used in proving the finite-\(p\) lemma?
- Why must the exceptional null sets for the \(L^\infty\) increment bounds be combined before defining the pointwise limit?
- How does the proof establish that the finite-\(p\) limit belongs to \(L^p(X)\), rather than merely being measurable?
- Why does convergence of the selected subsequence, by itself, not yet prove convergence of the original Cauchy sequence?