From Finite Linearity to Infinite Sums
The previous tutorial established that an integrable function is absolutely integrable: its absolute value has finite integral. This condition is important when sums are involved. Finite linearity, established in the earlier tutorial “Integrable Functions,” allows a finite sum to pass through the integral. An infinite sum requires additional control: the partial sums must not accumulate an uncontrolled amount of absolute value.
Throughout, \((X,\mathcal{F},\mu)\) is a measure space. A real-valued measurable function \(f\) is integrable when \(\int_X |f|\,d\mu<\infty\). We use the Integral of a Nonnegative Series theorem from earlier in this course: for nonnegative measurable functions, the integral of their pointwise sum is the sum of their integrals. We also use the Absolute-Value Bound for an Integrable Function and finite linearity, without reproving them here.
The condition is stronger than requiring each \(f_n\) to be integrable. It is also different from requiring the numerical series \(\sum_n\int_X f_n\,d\mu\) to converge: signed terms can cancel even when their absolute contributions are too large to control the corresponding functions.
Linearity for Absolutely Summable Series
For an absolutely summable sequence, the pointwise series need not converge at every point. The theorem below defines its sum to be zero on the exceptional set where absolute convergence fails. That set is null, so this convention does not affect the integral.
Proof. Set \(G_N=\sum_{n=1}^N|f_n|\), and let \(G=\lim_{N\to\infty}G_N=\sum_{n=1}^{\infty}|f_n|\), allowing the value \(+\infty\). Each \(G_N\) is measurable, so \(G\) is measurable. The functions \(G_N\) increase to \(G\). By the Monotone Convergence Theorem and finite additivity of the nonnegative integral, $$ \int_X G\,d\mu = \lim_{N\to\infty}\int_X G_N\,d\mu = \lim_{N\to\infty}\sum_{n=1}^N\int_X |f_n|\,d\mu = \sum_{n=1}^{\infty}\int_X |f_n|\,d\mu < \infty. $$ Consequently \(G<\infty\) almost everywhere. Indeed, if \(A=\{x:G(x)=\infty\}\) had positive measure, then \(G\geq m\mathbf{1}_A\) for every positive integer \(m\). Monotonicity would give \(\int_XG\,d\mu\geq m\mu(A)\) for every \(m\), which contradicts finiteness whether \(\mu(A)\) is finite and positive or infinite.
Let \(C=\{x:G(x)<\infty\}\), a measurable set. On \(C\), the series \(\sum_n f_n(x)\) converges absolutely. Its partial sums \(S_N=\sum_{n=1}^N f_n\) are measurable. The function equal to \(\mathbf{1}_C S_N\) on \(C\) and zero off \(C\) is measurable, and these functions converge pointwise to \(F\). Thus \(F\) is measurable. Moreover, \(|F|\leq G\) on \(C\), and \(F=0\) on \(X\setminus C\). It follows that \(|F|\leq G\) everywhere, so \(\int_X|F|\,d\mu\leq\int_XG\,d\mu<\infty\). Hence \(F\) is integrable.
It remains to identify its integral. For \(N\geq1\), set \(T_N=\sum_{n=N+1}^{\infty}|f_n|\). The Integral of a Nonnegative Series theorem gives $$ \int_X T_N\,d\mu = \sum_{n=N+1}^{\infty}\int_X|f_n|\,d\mu \longrightarrow 0. $$ On \(C\), absolute convergence implies \(|F-S_N|\leq T_N\). Since \(X\setminus C\) is null, this inequality holds almost everywhere. The monotonicity of the integral also gives the corresponding integral inequality for functions ordered almost everywhere: the part violating the order is supported on a null set and has integral zero. Therefore $$ \int_X|F-S_N|\,d\mu\leq\int_XT_N\,d\mu\longrightarrow0. $$ Both \(F\) and \(S_N\) are integrable. The Absolute-Value Bound and finite linearity yield $$ \left|\int_XF\,d\mu-\int_XS_N\,d\mu\right| = \left|\int_X(F-S_N)\,d\mu\right| \leq\int_X|F-S_N|\,d\mu \longrightarrow0. $$ By finite linearity, \(\int_XS_N\,d\mu=\sum_{n=1}^N\int_Xf_n\,d\mu\). Taking the limit proves the claimed equality. The numerical series on the right converges absolutely, since \(\left|\int_Xf_n\,d\mu\right|\leq\int_X|f_n|\,d\mu\). This completes the proof.
Worked Example: A Geometric Series of Constant Functions
Let \(X=[0,1]\) with Lebesgue measure, and define \(f_n(x)=2^{-n}\) for \(n\geq1\). Each \(f_n\) is integrable, and $$ \int_0^1|f_n(x)|\,dx=2^{-n}\lambda([0,1])=2^{-n}. $$ Therefore $$ \sum_{n=1}^{\infty}\int_0^1|f_n(x)|\,dx = \sum_{n=1}^{\infty}2^{-n} = 1<\infty. $$ At every \(x\in[0,1]\), the sum is \(F(x)=\sum_{n=1}^{\infty}2^{-n}=1\). The theorem gives $$ \int_0^1F(x)\,dx=1 = \sum_{n=1}^{\infty}2^{-n} = \sum_{n=1}^{\infty}\int_0^1f_n(x)\,dx. $$ Here the pointwise sum and the integral can both be computed directly; the theorem explains why the same equality holds for much less elementary sequences.
Worked Example: An Alternating Series with Absolute Control
Again take \(X=[0,1]\) with Lebesgue measure, and let \(f_n(x)=(-1)^{n+1}2^{-n}\). For every \(n\), $$ \int_0^1|f_n(x)|\,dx=2^{-n}, \qquad \sum_{n=1}^{\infty}\int_0^1|f_n(x)|\,dx=1. $$ Thus the series is absolutely summable in integral. Its pointwise sum is the constant function $$ F(x)=\sum_{n=1}^{\infty}(-1)^{n+1}2^{-n} =\frac{1/2}{1+1/2} =\frac13. $$ On the other hand, the integrals of its terms are \(\int_0^1f_n(x)\,dx=(-1)^{n+1}2^{-n}\), and their sum is also \(1/3\). Therefore $$ \int_0^1F(x)\,dx=\frac13 = \sum_{n=1}^{\infty}\int_0^1f_n(x)\,dx. $$ The absolute-summability hypothesis holds even though the terms change sign; it is what makes their cancellation safe to pass through the integral.
Decomposing an Integral Across a Countable Partition
A countable partition is a family of pairwise disjoint measurable sets whose union is the whole space. The next theorem says that an integrable function’s integral is the sum of its integrals over those pieces. It is a useful application of countable linearity: the pieces may be treated as the terms \(f\mathbf{1}_{E_n}\).
Proof. Define \(f_n=f\mathbf{1}_{E_n}\). Each \(f_n\) is measurable and satisfies \(|f_n|\leq|f|\), so it is integrable. Since the sets \(E_n\) partition \(X\), the nonnegative functions \(|f|\mathbf{1}_{E_n}\) sum pointwise to \(|f|\). The Integral of a Nonnegative Series theorem gives $$ \sum_{n=1}^{\infty}\int_X|f_n|\,d\mu = \sum_{n=1}^{\infty}\int_X|f|\mathbf{1}_{E_n}\,d\mu = \int_X|f|\,d\mu < \infty. $$ Thus \((f_n)\) is absolutely summable in integral. At every point \(x\), exactly one indicator \(\mathbf{1}_{E_n}(x)\) equals one, so \(\sum_n f_n(x)=f(x)\). Applying countable linearity gives $$ \int_X f\,d\mu = \sum_{n=1}^{\infty}\int_X f_n\,d\mu = \sum_{n=1}^{\infty}\int_{E_n}f\,d\mu. $$ Finally, the absolute convergence follows from $$ \sum_{n=1}^{\infty}\left|\int_{E_n}f\,d\mu\right| \leq \sum_{n=1}^{\infty}\int_{E_n}|f|\,d\mu = \int_X|f|\,d\mu<\infty. $$ This proves both claims.
Worked Example: Splitting an Integral on the Half-Line
Let \(X=[0,\infty)\) with Lebesgue measure, take \(f(x)=(1+x)^{-2}\), and partition \(X\) into \(E_n=[n,n+1)\) for \(n=0,1,2,\ldots\). The function is nonnegative, and direct integration gives $$ \int_0^\infty(1+x)^{-2}\,dx = \lim_{R\to\infty}\left[-\frac{1}{1+x}\right]_0^R = 1. $$ For each \(n\geq0\), $$ \int_{E_n}f(x)\,dx = \left[-\frac{1}{1+x}\right]_n^{n+1} = \frac{1}{n+1}-\frac{1}{n+2}. $$ The partial sum through \(n=N\) telescopes: $$ \sum_{n=0}^N\int_{E_n}f(x)\,dx = 1-\frac{1}{N+2} \longrightarrow1. $$ This agrees with the integral over the whole half-line and illustrates how the partition formula turns a global integral into a sum of local contributions.
Why Absolute Control Matters
Countable linearity is not simply finite linearity with an infinite number of terms substituted in. The proof needs the tails to become small in integral. The condition \(\sum_n\int_X|f_n|\,d\mu<\infty\) provides exactly that control: it makes the integral of the absolute tail tend to zero, and hence makes the integrals of the partial sums converge to the integral of the sum.
In particular, convergence of the numerical series \(\sum_n\int_X f_n\,d\mu\) alone is not the theorem’s hypothesis. Such convergence can result from cancellation between positive and negative terms, without controlling the pointwise sum or its integral. Absolute summability rules out this gap. For a countable partition, the pieces \(f\mathbf{1}_{E_n}\) automatically satisfy the needed condition because their absolute integrals add to \(\int_X|f|\,d\mu\).
Check Your Understanding
Use the series theorem and the partition formula to answer these questions.
- Why does \(\sum_n\int_X|f_n|\,d\mu<\infty\) imply that \(\sum_n|f_n(x)|\) is finite almost everywhere?
- In the proof of countable linearity, what estimate shows that the integrals of the partial sums approach the integral of the sum?
- Why does convergence of \(\sum_n\int_X f_n\,d\mu\) alone not supply the absolute control used in the theorem?
- How does a countable measurable partition produce a sequence to which countable linearity applies?
- For \(f_n(x)=(-1)^{n+1}2^{-n}\) on \([0,1]\), compute \(\sum_n\int_0^1|f_n|\,dx\) and \(\int_0^1\sum_n f_n(x)\,dx\).