Applications of Monotone Convergence

From the Theorem to Useful Calculations

The Monotone Convergence Theorem, proved in the previous tutorial, turns pointwise increase into convergence of integrals. Its applications include approximating a function on larger and larger parts of a space and integrating a sum of nonnegative functions one term at a time. Both applications are useful because they replace an infinite object with finite stages whose integrals can often be computed directly.

There is one point to handle before exchanging an integral with a series: the finite partial sum must have an integral equal to the sum of the individual integrals. The earlier course established this for nonnegative simple functions. We first extend finite additivity to general nonnegative measurable functions, using simple approximations and the Monotone Convergence Theorem. We can then pass from finite sums to infinite sums.

Finite Additivity for Nonnegative Functions

Theorem (Finite Additivity for Nonnegative Integrals): If $f,g:X\to[0,\infty]$ are measurable on a measure space, then $$ \int_X(f+g)\,d\mu=\int_X f\,d\mu+\int_X g\,d\mu, $$ where the equality is in $[0,\infty]$.

Proof. We use increasing simple approximations for $f$ and $g$. Such approximations also exist when a function may take the value $+\infty$: truncate the function at $n$, then round the truncated values down to the nearest multiple of $2^{-n}$. In particular, we can choose nonnegative measurable simple functions $s_n$ and $t_n$ such that $s_n\uparrow f$ and $t_n\uparrow g$. The truncation and rounding errors tend to zero at each point where the function is finite; at a point where it is infinite, the truncation levels tend to infinity. Thus the approximations converge pointwise in both cases.

The functions $s_n+t_n$ are nonnegative measurable simple functions, and $s_n+t_n\uparrow f+g$ pointwise. The finite additivity result for simple-function integrals gives $$ \int_X(s_n+t_n)\,d\mu =\int_Xs_n\,d\mu+\int_Xt_n\,d\mu. $$ Apply the Monotone Convergence Theorem to $s_n$, $t_n$, and $s_n+t_n$. The left side tends to $\int_X(f+g)\,d\mu$; the two terms on the right tend to $\int_X f\,d\mu$ and $\int_X g\,d\mu$. For nonnegative increasing sequences, the limit of the sums is the sum of the limits, including when either limit is $+\infty$. Therefore $$ \int_X(f+g)\,d\mu =\int_X f\,d\mu+\int_X g\,d\mu. $$ This proves the theorem.

Applying this result repeatedly shows that the integral of any finite sum of nonnegative measurable functions is the sum of their integrals. We will use precisely that finite-stage identity when passing to a countable sum. No subtraction of infinite quantities is involved.

Integrating a Nonnegative Series

Theorem (Integral of a Nonnegative Series): Let $f_k:X\to[0,\infty]$ be measurable for each positive integer $k$, and define $$ f(x)=\sum_{k=1}^{\infty}f_k(x) =\lim_{n\to\infty}\sum_{k=1}^{n}f_k(x). $$ Then $f$ is measurable and $$ \int_X\left(\sum_{k=1}^{\infty}f_k\right)d\mu =\sum_{k=1}^{\infty}\int_X f_k\,d\mu, $$ with both sides interpreted in $[0,\infty]$.

Proof. Set $S_n=\sum_{k=1}^{n}f_k$. Each $S_n$ is measurable, and $S_n\leq S_{n+1}$ because $f_{n+1}\geq0$. The pointwise limit $f=\lim_n S_n$ is measurable by the Pointwise Limits of Measurable Functions theorem. The Monotone Convergence Theorem now gives $$ \int_X f\,d\mu=\lim_{n\to\infty}\int_X S_n\,d\mu. $$ By finite additivity for nonnegative integrals, $$ \int_X S_n\,d\mu=\sum_{k=1}^{n}\int_X f_k\,d\mu. $$ The right side increases to the nonnegative series $\sum_{k=1}^{\infty}\int_X f_k\,d\mu$, by the definition of a series of nonnegative terms. Combining these equalities proves the result, including the case where the limit is infinite.

The nonnegativity assumption is essential to this argument: it ensures the partial sums increase. The theorem makes no claim that an arbitrary series of signed functions can be integrated term by term. For signed terms, cancellation can affect convergence, and additional hypotheses are needed.

Worked Example: A Geometric Series of Constant Functions

Let $X=[0,1]$ with Lebesgue measure, and for $k\geq1$ define $f_k(x)=2^{-k}$ for every $x\in X$. Each function is nonnegative and measurable. For every $x$, $$ \sum_{k=1}^{\infty}f_k(x)=\sum_{k=1}^{\infty}2^{-k}=1. $$ Since $\lambda([0,1])=1$, the integral of each constant function is $$ \int_{[0,1]}f_k\,d\lambda=2^{-k}. $$ The theorem gives $$ \int_{[0,1]}\left(\sum_{k=1}^{\infty}f_k\right)d\lambda =\sum_{k=1}^{\infty}2^{-k}=1. $$ Directly, the left side is $\int_{[0,1]}1\,d\lambda=1$, agreeing with the series calculation. This example is simple, but it displays the finite-stage idea: the integrals of the partial sums are the partial geometric sums, which increase to $1$.

Worked Example: Disjoint Intervals and a Divergent Integral

On $\mathbb{R}$ with Lebesgue measure, let $A_k=[k,k+1)$ for each $k\geq1$, and set $f_k=\mathbf{1}_{A_k}$. The intervals are pairwise disjoint, and each has measure $1$, so the Integral of an Indicator theorem gives $$ \int_{\mathbb{R}}f_k\,d\lambda=\lambda(A_k)=1. $$ At each point $x$, at most one of the functions $f_k(x)$ is nonzero. Their sum is therefore the indicator of $\bigcup_{k=1}^{\infty}A_k=[1,\infty)$. The series theorem yields $$ \int_{\mathbb{R}}\mathbf{1}_{[1,\infty)}\,d\lambda =\sum_{k=1}^{\infty}1 =+\infty. $$ The equality does not require the union to have finite measure. Both sides are well-defined as nonnegative extended values, and here both are infinite.

Increasing Domains

A second useful form of monotone convergence concerns restriction to larger and larger measurable sets. For a measurable set $E$, write $\int_E f\,d\mu$ for $\int_X f\mathbf{1}_E\,d\mu$, where $f$ is nonnegative measurable. If the sets increase, their indicators increase pointwise, so the restricted functions form an increasing sequence.

Theorem (Approximation on Increasing Domains): Let $f:X\to[0,\infty]$ be measurable, and let $E_1\subseteq E_2\subseteq\cdots$ be measurable. If $E=\bigcup_{n=1}^{\infty}E_n$, then $$ \lim_{n\to\infty}\int_{E_n}f\,d\mu=\int_E f\,d\mu. $$

Proof. For each $n$, define $g_n=f\mathbf{1}_{E_n}$. These functions are measurable and nonnegative. Since $E_n\subseteq E_{n+1}$, we have $g_n(x)\leq g_{n+1}(x)$ for every $x$. If $x\in E$, then $x\in E_N$ for some $N$, so $g_n(x)=f(x)$ for every $n\geq N$. If $x\notin E$, then $g_n(x)=0$ for every $n$. Thus $g_n$ increases pointwise to $f\mathbf{1}_E$. The Monotone Convergence Theorem gives $$ \lim_{n\to\infty}\int_X g_n\,d\mu =\int_X f\mathbf{1}_E\,d\mu. $$ By the definition of integration over a measurable set, this is the stated equality.

This result requires neither $E$ nor the individual $E_n$ to have finite measure. It also does not require $f$ to be bounded. If the integral on $E$ is infinite, the integrals on the increasing domains tend to $+\infty$.

Worked Example: Integrating over Larger and Larger Intervals

Take $X=\mathbb{R}$, $f(x)=1$, and $E_n=[-n,n]$. The sets increase and their union is $\mathbb{R}$. Each interval has measure $2n$, and therefore $$ \int_{E_n}1\,d\lambda=\lambda([-n,n])=2n. $$ As $n\to\infty$, these integrals tend to $+\infty$. The increasing-domains theorem identifies their limit with $$ \int_{\mathbb{R}}1\,d\lambda=+\infty. $$ This is a useful reminder that convergence in the theorem is convergence in the extended nonnegative reals; it need not have a finite limit.

Disjoint Pieces and Practical Checks

The series theorem also gives a convenient rule for splitting an integral across countably many disjoint measurable pieces. If $A_1,A_2,\ldots$ are pairwise disjoint and $A=\bigcup_{k=1}^{\infty}A_k$, then pointwise $$ f\mathbf{1}_A=\sum_{k=1}^{\infty}f\mathbf{1}_{A_k}. $$ Indeed, a point outside $A$ contributes zero to every term, while a point in $A$ belongs to exactly one of the sets. Applying the theorem to the functions $f\mathbf{1}_{A_k}$ gives $$ \int_A f\,d\mu=\sum_{k=1}^{\infty}\int_{A_k}f\,d\mu. $$ This extends the finite additivity of the integral across disjoint pieces to a countable decomposition, without requiring a finite total integral.

When using these applications, check three points. The functions or partial sums must be measurable; the sequence to which Monotone Convergence is applied must be nonnegative and increasing; and the limiting function must be the one whose integral is being computed. For a nonnegative series, increase follows from nonnegative terms. For increasing domains, it follows from inclusion of the sets. If any of these checks fails, the theorem does not by itself justify the limiting step.

Key takeaway: Monotone convergence permits limits over increasing domains and allows a nonnegative series to pass through the integral. The resulting equalities remain valid when their common value is $+\infty$.

Check Your Understanding

Use the hypotheses and conclusions of the applications above to answer these questions.

Why do the partial sums of a series of nonnegative measurable functions increase pointwise?
Which previously established theorem justifies passing from the increasing partial sums to the integral of their pointwise limit?
For increasing measurable sets $E_n$, what is the pointwise limit of $f\mathbf{1}_{E_n}$ when $E=\bigcup_n E_n$?
Why does the disjointness of the sets $A_k$ give the pointwise identity $f\mathbf{1}_{\cup_k A_k}=\sum_k f\mathbf{1}_{A_k}$?
Does the integral-of-a-series theorem require the sum of the individual integrals to be finite? Explain.