Why Completeness Matters
A normed space is complete if every Cauchy sequence converges in its norm to an element of the same space. In an \(L^p\) space, this says more than that a sequence of functions has a pointwise limit: the limit must be measurable, belong to \(L^p\), and approximate the sequence in the \(L^p\) norm. The central construction is to pass from a Cauchy sequence to a subsequence whose successive differences have summable norms. Those differences can then be added to construct a limit.
Let \((X,\mathcal{M},\mu)\) be any measure space; it need not be finite, sigma-finite, or complete. Functions may be real-valued or complex-valued, and \(L^p(\mu)\) consists of equivalence classes modulo equality \(\mu\)-a.e. For \(1\le p<\infty\), membership means that \(f\) is measurable and \(\int_X |f|^p\,d\mu<\infty\), with norm \(\|f\|_p=(\int_X|f|^p\,d\mu)^{1/p}\). The space \(L^\infty(\mu)\) consists of measurable functions with finite essential supremum norm \(\|f\|_\infty\). All norm statements concern these equivalence classes.
We use Minkowski’s inequality, which states that \(\|u+v\|_p\le\|u\|_p+\|v\|_p\) for \(1\le p\le\infty\) when \(u,v\in L^p(\mu)\). We also use the Monotone Convergence Theorem for nonnegative measurable functions. The measure space assumptions for both results are just those stated above.
The Riesz–Fischer Theorem
Proof. Let \((f_n)\) be Cauchy in \(L^p(\mu)\). For each \(j\ge1\), the Cauchy property gives an index \(N_j\) such that \(\|f_n-f_m\|_p<2^{-j}\) whenever \(n,m\ge N_j\). Choose strictly increasing indices \(n_j\) with \(n_j\ge N_j\). Then \(n_{j+1}\ge N_j\) as well, so \(\|f_{n_{j+1}}-f_{n_j}\|_p<2^{-j}\). Choose measurable representatives and write \(g_j=f_{n_{j+1}}-f_{n_j}\). Set \(a_j=\|g_j\|_p\), so \(\sum_{j=1}^\infty a_j\le\sum_{j=1}^\infty2^{-j}=1\).
First suppose \(1\le p<\infty\). For \(N\ge1\), define the nonnegative measurable function \(H_N=\sum_{j=1}^N|g_j|\). Minkowski’s inequality gives \(\|H_N\|_p\le\sum_{j=1}^N a_j\le1\). The functions \(H_N^p\) increase pointwise to \(H^p\), where \(H=\sum_{j=1}^\infty|g_j|\) is allowed initially to take the value \(+\infty\). By the Monotone Convergence Theorem, \(\int_X H^p\,d\mu=\lim_N\int_X H_N^p\,d\mu\le1\). Thus \(H\) is finite \(\mu\)-a.e. Define the measurable set \(S=\{x:H(x)<\infty\}\), which has null complement. On \(S\), the telescoping sums \(f_{n_1}+\sum_{j=1}^{N}g_j=f_{n_{N+1}}\) converge pointwise to a finite limit. Define \(f\) to be this limit on \(S\) and \(0\) on \(S^c\). This is measurable: the pointwise limit on the measurable set \(S\), extended by zero on \(S^c\), is measurable. Completeness of the measure space is not needed.
For each \(j\), on \(S\) we have \(|f-f_{n_j}|\le\sum_{k=j}^\infty|g_k|\), and the inequality holds a.e. on \(X\). Applying Minkowski to finite sums and then Monotone Convergence to their \(p\)-th powers gives \(\|f-f_{n_j}\|_p\le\sum_{k=j}^\infty a_k\le\sum_{k=j}^\infty2^{-k}=2^{1-j}\). In particular, \(f-f_{n_1}\in L^p(\mu)\), so \(f\in L^p(\mu)\). The same estimate shows \(\|f-f_{n_j}\|_p\to0\).
Now suppose \(p=\infty\). For every \(j\), the definition of essential supremum gives a measurable null set \(Z_j\) outside which \(|g_j|\le a_j\); if \(a_j=0\), this inequality still holds a.e. Let \(S=X\setminus\bigcup_{j=1}^\infty Z_j\). This is measurable and has null complement. For \(x\in S\), every finite telescoping sum satisfies \(|f_{n_{N+1}}(x)-f_{n_1}(x)|\le\sum_{j=1}^N a_j\le1\). The increments are absolutely summable at each \(x\in S\), so the subsequence converges there to a finite limit. Define \(f\) as that limit on \(S\) and zero on \(S^c\); it is measurable. Since \(f_{n_1}\in L^\infty(\mu)\) and \(|f-f_{n_1}|\le1\) on \(S\), \(f\in L^\infty(\mu)\). The tail bound gives \(\|f-f_{n_j}\|_\infty\le\sum_{k=j}^\infty a_k\le2^{1-j}\), so the subsequence converges to \(f\) in norm.
It remains, in either case, to pass from the subsequence to the original sequence. Given \(\varepsilon>0\), choose \(N\) such that \(\|f_n-f_m\|_p<\varepsilon/2\) for \(n,m\ge N\). Choose \(j\) with \(n_j\ge N\) and \(\|f_{n_j}-f\|_p<\varepsilon/2\). For every \(n\ge N\), the triangle inequality yields \(\|f_n-f\|_p\le\|f_n-f_{n_j}\|_p+\|f_{n_j}-f\|_p<\varepsilon\). Thus the entire sequence converges to \(f\) in \(L^p(\mu)\). \(\square\)
Worked Consequences of the Construction
Worked Example: A Geometric Sequence on a Finite-Measure Set
On \([0,1]\) with Lebesgue measure, let \(f_n=(1-2^{-n})\chi_{[0,1]}\). Compute its limit and its distance from that limit in \(L^p\), for each \(1\le p\le\infty\).
At every \(x\in[0,1]\), the scalar coefficient tends to \(1\), so \(f_n\to f=\chi_{[0,1]}\) pointwise. For finite \(p\), the difference is \(f-f_n=2^{-n}\chi_{[0,1]}\). Since \(m([0,1])=1\), \(\|f-f_n\|_p=(\int_0^1(2^{-n})^p\,dm)^{1/p}=(2^{-np}\cdot1)^{1/p}=2^{-n}\). For \(p=\infty\), its essential supremum is also \(2^{-n}\). Thus the norm error tends to zero for every \(p\) in the theorem. Check: at \(n=2\), the error is \(2^{-2}=1/4\), both by the displayed integral for finite \(p\) and by the essential supremum for \(p=\infty\).
Worked Example: A Limit Built from Disjoint Intervals
On \(\mathbb{R}\) with Lebesgue measure, define \(f_n=\sum_{k=1}^n3^{-k}\chi_{[k,k+1)}\). Find the limit and the exact norm of its tail for \(1\le p\le\infty\).
The intervals \([k,k+1)\) are pairwise disjoint and each has measure \(1\). Hence the pointwise limit is \(f=\sum_{k=1}^\infty3^{-k}\chi_{[k,k+1)}\). If \(p<\infty\), disjointness gives \(\|f-f_n\|_p^p=\sum_{k=n+1}^\infty 3^{-kp}\). This geometric series has first term \(3^{-(n+1)p}\) and ratio \(3^{-p}\), which is less than \(1\). Therefore \(\|f-f_n\|_p^p=3^{-(n+1)p}/(1-3^{-p})\), and \(\|f-f_n\|_p=3^{-(n+1)}(1-3^{-p})^{-1/p}\). For \(p=\infty\), the tail values are \(3^{-(n+1)},3^{-(n+2)},\ldots\), so \(\|f-f_n\|_\infty=3^{-(n+1)}\). Both formulas tend to zero, and in particular \(f\in L^p(\mathbb{R})\) in each stated case. Check: for \(p=1,n=1\), the tail integral is \(\sum_{k=2}^\infty3^{-k}=(1/9)/(1-1/3)=1/6\), matching \(3^{-2}/(1-1/3)=1/6\).
Why Pointwise Convergence Is Not Enough
Worked Example: Moving Spikes Are Not Cauchy in Lp
For \(1\le p<\infty\), on \(\mathbb{R}\) with Lebesgue measure let \(u_n=n^{1/p}\chi_{(0,1/n)}\). Show that \(u_n\to0\) pointwise but the sequence is not Cauchy in \(L^p\).
Fix \(x>0\). Once \(n\ge1/x\), \(x\notin(0,1/n)\), so \(u_n(x)=0\); also \(u_n(0)=0\). Thus \(u_n(x)\to0\) everywhere. Yet \(\|u_n\|_p^p=n\cdot m((0,1/n))=n(1/n)=1\). More directly, take \(m=2n\). On \((1/(2n),1/n)\), which has measure \(1/(2n)\), \(u_n=n^{1/p}\) and \(u_{2n}=0\). Consequently \(\|u_n-u_{2n}\|_p^p\ge n\cdot(1/(2n))=1/2\), so \(\|u_n-u_{2n}\|_p\ge(1/2)^{1/p}>0\) for every \(n\). The sequence is not Cauchy. Check: at \(p=1\), the lower bound is \(1/2\); at \(p=2\), it is \(1/\sqrt{2}\). This example does not contradict completeness: completeness applies to norm-Cauchy sequences, not to all sequences that converge pointwise.
A common mistake is to infer \(L^p\) convergence merely from pointwise convergence, or to construct a pointwise limit without checking it belongs to \(L^p\). The summable-increment argument addresses both issues: for finite \(p\), it bounds the \(L^p\) norm of the sum of absolute increments; for \(p=\infty\), it gives a uniform bound outside one null set. In both cases the tail of the summable norms controls the distance to the limit. The final Cauchy estimate is essential too: it is what extends convergence from the carefully chosen subsequence to the original sequence.
Check Your Understanding
Use the theorem and its proof to check each claim, keeping track of which norm and which type of convergence are involved.
- Why must the subsequence indices be chosen so that each successive pair lies beyond the Cauchy threshold for its corresponding \(2^{-j}\)?
- For finite \(p\), what does Monotone Convergence apply to in the proof that the sum of absolute increments belongs to \(L^p\)?
- Why is a countable union of null sets sufficient to obtain one set on which all \(L^\infty\) increment bounds hold simultaneously?
- For \(f_n=(1-3^{-n})\chi_{[0,1]}\) on \([0,1]\), compute \(\|f-f_n\|_p\) for finite \(p\) and \(\|f-f_n\|_\infty\).
- Why does the moving-spike sequence in the example fail to contradict completeness, even though it converges pointwise?