What Makes \(L^1\) a Space?
The previous tutorial established Hölder’s inequality, Minkowski’s inequality, and the norm properties of \(L^p\). For \(p=1\), the norm measures the total absolute size of a function: \(\|f\|_1=\int_X|f|\,d\mu\). The phrase “the \(L^1\) space” refers not just to a collection of functions with finite norm, but to a vector space in which functions equal almost everywhere are treated as the same element.
This identification matters because the integral cannot distinguish functions that differ only on a set of measure zero. The central structural result for \(L^1\) is completeness: every Cauchy sequence in its norm has a limit in the space. We will prove this by selecting a subsequence whose successive differences have summable \(L^1\) norms. The resulting series of differences converges almost everywhere and controls the norm error.
The definition is independent of the representative: if \(f=g\) almost everywhere, then \(|f|=|g|\) almost everywhere and their integrals agree. Earlier in the course, the Invariance of the Integral Under Almost-Everywhere Equality theorem established this principle for integrable functions. The zero-norm criterion says that \(\|f\|_1=0\) exactly when \(f=0\) almost everywhere, which is why the quotient by almost-everywhere equality is necessary for a genuine norm.
Linearity of \(L^1\) follows from the Linear Closure of \(L^p\) with \(p=1\), and the norm axioms follow from the results in the previous tutorials. Thus the new issue is not whether \(L^1\) is a normed vector space, but whether its norm has enough limits to make it complete.
Worked Example: The Norm of an Indicator Function
On \(\mathbb{R}\) with Lebesgue measure, let \(A=[-2,1]\) and \(f=\mathbf{1}_A\). The Integral of an Indicator theorem gives $$ \|f\|_1=\int_{\mathbb{R}}\mathbf{1}_A\,d\lambda =\lambda(A)=1-(-2)=3. $$ So the indicator belongs to \(L^1(\mathbb{R},\lambda)\). If its value is changed at either endpoint, its equivalence class and \(L^1\) norm remain unchanged, since a finite set has Lebesgue measure zero.
Worked Example: A Singular Function in \(L^1\)
Consider \(f(x)=x^{-2/3}\) on \((0,1)\) with Lebesgue measure. This function is nonnegative, so its \(L^1\) norm is its integral. For \(0<a<1\), $$ \int_a^1 x^{-2/3}\,dx =3\left(1-a^{1/3}\right). $$ As \(a\) decreases to zero, these integrals increase to \(3\). Hence \(\|f\|_1=3\), and \(f\in L^1(0,1)\), even though it is unbounded near zero. By contrast, \(\int_a^1 x^{-1}\,dx=-\log a\) tends to infinity as \(a\) decreases to zero, so \(x^{-1}\) does not belong to \(L^1(0,1)\). Unboundedness alone does not decide \(L^1\) membership; the total absolute integral does.
A Summable-Differences Principle
The key construction for proving completeness is a series whose terms are differences between selected functions. If the sum of the \(L^1\) norms of those differences is finite, then the pointwise sum of their absolute values is integrable. This follows from the Integral of a Nonnegative Series theorem, established earlier in the course.
Proof. Choose measurable representatives for the \(u_k\), and define the nonnegative measurable function $$ H(x)=\sum_{k=1}^{\infty}|u_k(x)|. $$ By the Integral of a Nonnegative Series theorem, $$ \int_X H\,d\mu =\sum_{k=1}^{\infty}\int_X|u_k|\,d\mu =\sum_{k=1}^{\infty}\|u_k\|_1<\infty. $$ A nonnegative function with finite integral is finite almost everywhere: if \(H=\infty\) on a set of positive measure, its integral would be infinite. Thus the measurable set \(E=\{x:H(x)<\infty\}\) has complement of measure zero. On \(E\), the numerical series \(\sum_k u_k(x)\) converges absolutely. Define \(u(x)\) to be this sum on \(E\), and set \(u(x)=0\) on \(X\setminus E\). The partial sums are measurable and converge on \(E\), so this definition gives a measurable function.
On \(E\), the triangle inequality for convergent numerical series gives \(|u|\leq H\). Since \(u=0\) off \(E\), it follows that \(\int_X|u|\,d\mu\leq\int_XH\,d\mu<\infty\), so \(u\in L^1\). Let \(s_N=\sum_{k=1}^N u_k\). On \(E\), $$ |u-s_N|\leq\sum_{k=N+1}^{\infty}|u_k|. $$ The complement of \(E\) is null, so integrating and applying the Integral of a Nonnegative Series theorem again yields $$ \|u-s_N\|_1 \leq\sum_{k=N+1}^{\infty}\|u_k\|_1. $$ The right-hand side tends to zero because the series of norms converges. Therefore \(s_N\to u\) in \(L^1\), as claimed.
Worked Example: A Series on Disjoint Intervals
On \((0,1)\), set \(E_k=(2^{-k},2^{-(k-1)}]\) and \(u_k=\mathbf{1}_{E_k}\) for each positive integer \(k\). The intervals are disjoint and have measure $$ \lambda(E_k)=2^{-(k-1)}-2^{-k}=2^{-k}. $$ Consequently, \(\|u_k\|_1=2^{-k}\), and \(\sum_{k=1}^{\infty}\|u_k\|_1=1\). Their sum is \(1\) at every point of \((0,1)\), since the intervals partition \((0,1)\). The theorem therefore gives \(L^1\) convergence of the partial sums to \(1\). Directly, the error after \(N\) terms is the indicator of \((0,2^{-N}]\), up to endpoints, and its norm is \(2^{-N}\), which tends to zero.
Completeness of \(L^1\)
A normed space is complete if every Cauchy sequence in its norm converges to an element of the space. For \(L^1\), we do not initially know a limit function exists. We first extract a subsequence with summable successive differences, then use the preceding theorem to construct its limit.
Proof. Let \((F_n)\) be a Cauchy sequence in \(L^1\), where each \(F_n\) is an equivalence class. By the Cauchy property, for each positive integer \(j\) there is an index \(N_j\) such that $$ \|F_r-F_s\|_1<2^{-j} \quad\text{whenever }r,s\geq N_j. $$ Choose strictly increasing indices \(n_j\) with \(n_j\geq N_j\). Then $$ \|F_{n_{j+1}}-F_{n_j}\|_1<2^{-j}. $$ Choose measurable representatives \(f_{n_j}\) of these classes and define \(u_j=f_{n_{j+1}}-f_{n_j}\). The summable-differences estimate gives $$ \sum_{j=1}^{\infty}\|u_j\|_1 \leq\sum_{j=1}^{\infty}2^{-j}=1. $$ By the Absolute Convergence of a Series in \(L^1\) theorem, the series \(\sum_j u_j\) has a sum \(u\in L^1\), and its partial sums converge to \(u\) in \(L^1\).
For every \(m\geq1\), telescoping gives $$ f_{n_m}=f_{n_1}+\sum_{j=1}^{m-1}u_j. $$ Since \(f_{n_1}\in L^1\) and the partial sums of the series converge in \(L^1\) to \(u\), the selected subsequence \(F_{n_m}\) converges in \(L^1\) to the class \(F\) represented by \(f_{n_1}+u\). In particular, \(F\in L^1\).
It remains to show that the full sequence converges to \(F\), not just the selected subsequence. Given \(\varepsilon>0\), choose \(N\) so that \(\|F_r-F_s\|_1<\varepsilon/2\) whenever \(r,s\geq N\). Because \(F_{n_m}\to F\), choose \(m\) large enough that \(n_m\geq N\) and \(\|F_{n_m}-F\|_1<\varepsilon/2\). For every \(n\geq N\), the triangle inequality gives $$ \|F_n-F\|_1 \leq\|F_n-F_{n_m}\|_1+\|F_{n_m}-F\|_1 <\varepsilon. $$ Thus \(F_n\to F\) in \(L^1\), proving completeness.
Reading \(L^1\) Convergence Correctly
Convergence in \(L^1\) controls the integral of the absolute error, not the largest pointwise error. A sequence can converge in \(L^1\) while its functions remain far from the limit at some points. The next example makes this distinction explicit.
Worked Example: Tall, Narrow Functions
On \((0,1)\), let \(f_n(x)=\sqrt{n}\,\mathbf{1}_{(0,1/n)}(x)\). Then $$ \|f_n\|_1 =\sqrt{n}\,\lambda((0,1/n)) =\frac{1}{\sqrt{n}}\longrightarrow0. $$ Therefore \(f_n\) converges to zero in \(L^1\). For each fixed \(x\in(0,1)\), once \(n>1/x\) we have \(x>1/n\), so \(f_n(x)=0\); hence the sequence also converges pointwise to zero. But its essential supremum is \(\sqrt{n}\), which grows without bound. In particular, the convergence is not uniform: for every \(n\), points \(x\in(0,1/n)\) satisfy \(|f_n(x)|=\sqrt{n}\), rather than a quantity uniformly tending to zero.
Completeness is useful whenever approximations are controlled by a summable sequence of errors. One can first construct approximating functions whose successive \(L^1\) differences have a finite total, then obtain a genuine \(L^1\) limit without guessing its formula in advance. The Absolute Convergence of a Series in \(L^1\) theorem is one direct form of this principle.
A common pitfall is to confuse completeness with pointwise convergence. Completeness begins with a Cauchy condition in the norm; it does not claim that every pointwise-convergent sequence converges in \(L^1\). Conversely, as the tall, narrow example shows, \(L^1\) convergence is not uniform convergence. The metric being used matters: the \(L^1\) norm measures total integrated error.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Why are functions equal almost everywhere identified when defining \(L^1\)?
- What condition on a sequence \((u_k)\) ensures that its series converges in \(L^1\) by the Absolute Convergence of a Series in \(L^1\) theorem?
- In the completeness proof, why is a subsequence with summable successive differences useful?
- Can an unbounded function belong to \(L^1\)? Explain using an example from the tutorial.
- Why does convergence in \(L^1\) not imply uniform convergence, as illustrated by the tall, narrow functions?