Monotone Convergence on Counting Measure

Counting Measure Turns Integrals into Sums

On a countable set, integration with respect to counting measure is summation. This makes the Monotone Convergence Theorem a precise tool for passing from finite sums to an infinite sum, even when the answer is \(+\infty\). The central requirement is that the entries increase as the approximation index increases; pointwise convergence by itself does not justify the same passage.

Let \(I\) be a countable set, let \(\mathcal{P}(I)\) be its power-set sigma-algebra, and let \(\#\) be counting measure, defined by \(\#(E)=|E|\) when \(E\) is finite and \(\#(E)=+\infty\) when \(E\) is infinite. The measure space \((I,\mathcal{P}(I),\#)\) is sigma-finite when \(I\) is countable: it is a countable union of singletons, each of measure \(1\). It is finite exactly when \(I\) is finite. Every function on \(I\) is measurable. In this tutorial, array entries take values in \([0,+\infty]\).

To see why the integral identity holds, each finite sum \(\sum_{i\in F}f(i)\) is the integral of the function \(f\chi_F\), so it is at most \(\int_I f\,d\#\). Conversely, the integral of a nonnegative simple function is the supremum of its finite sums over finite subsets of \(I\). Since every simple function below \(f\) has each of its values bounded above by the corresponding value of \(f\), its integral is at most the supremum in the definition. Taking the supremum over simple functions below \(f\), as in the definition of the nonnegative integral, gives the reverse inequality.

Increasing Arrays and Their Sums

An array here has an approximation index \(r\in\mathbb{N}=\{1,2,\ldots\}\) and a countable-set index \(i\in I\). Its entries \(a_{r,i}\) are nonnegative, and “increasing” refers to the approximation index: \(a_{r,i}\le a_{r+1,i}\) for every \(r\) and \(i\). The countable-set index is summed over; it is not being ordered by size.

Proof. If \(I=\varnothing\), both sums are \(0\), so the identity holds. Otherwise, consider the measure space \((I,\mathcal{P}(I),\#)\). The functions \(f_r(i)=a_{r,i}\) are measurable, nonnegative, and increase pointwise to \(f(i)=a_i\). The Monotone Convergence Theorem, applied on this measure space, gives \(\int_I f\,d\#=\lim_{r\to\infty}\int_I f_r\,d\#\). By the counting-measure integral identity, the left side is \(\sum_{i\in I}a_i\) and each integral on the right is \(\sum_{i\in I}a_{r,i}\). This proves the assertion, including when the common value is infinite. \(\square\)

The sums in this theorem need not be finite at any stage. Nor does the theorem require \(I\) to be finite or the measure space to have finite measure: countability gives sigma-finiteness, but monotone convergence itself applies without a finiteness assumption.

Finite Partial Sums and Exhaustions

Suppose \(I\) is countably infinite and is enumerated as \(i_1,i_2,\ldots\). Let \(F_N=\{i_1,\ldots,i_N\}\). For any fixed \(r\), the finite sums \(\sum_{i\in F_N}a_{r,i}\) increase to \(\sum_{i\in I}a_{r,i}\). More generally, one can let the approximation index and the finite set grow together.

Proof. Write \(L=\sum_{i\in I}\lim_{r\to\infty}a_{r,i}\). For each \(N\), the finite sum \(\sum_{i\in F_N}a_{N,i}\) is at most \(L\), because \(a_{N,i}\le\lim_r a_{r,i}\) for every \(i\). The diagonal sums are increasing: \(F_N\subseteq F_{N+1}\), all entries are nonnegative, and \(a_{N,i}\le a_{N+1,i}\). Thus their limit exists and is at most \(L\). To prove the reverse inequality, fix any finite \(F\subseteq I\) and any \(r\). Since the sets \(F_N\) exhaust \(I\), there is an \(N_0\) such that \(F\subseteq F_N\) for all \(N\ge N_0\). For \(N\ge\max(N_0,r)\), monotonicity gives $$ \sum_{i\in F_N}a_{N,i}\ge\sum_{i\in F}a_{N,i}\ge\sum_{i\in F}a_{r,i}. $$ Consequently, the limit of the diagonal sums is at least \(\sum_{i\in F}a_{r,i}\) for every finite \(F\) and every \(r\). Taking the supremum over finite \(F\) gives a lower bound \(\sum_{i\in I}a_{r,i}\) for each \(r\). Taking the supremum over \(r\), the theorem gives the lower bound \(L\). Combined with the upper bound, this proves equality. If \(I\) is finite, take \(F_N=I\) for all sufficiently large \(N\); the assertion then follows directly from the theorem. \(\square\)

Worked Arrays

Why Pointwise Convergence Alone Is Not Enough

The increase condition is substantive. On \(\mathbb{N}\), define \(d_r(k)=r\) if \(k=r\), and \(d_r(k)=0\) otherwise. For each fixed \(k\), \(d_r(k)\) is nonzero at only the single index \(r=k\), so \(d_r(k)\to0\). Thus the pointwise limit has sum \(0\). However, \(\sum_{k=1}^{\infty}d_r(k)=r\) for every \(r\), so the sums tend to \(+\infty\), not \(0\). The sequence is not increasing: at the coordinate \(k=r\), the value drops from \(r\) at stage \(r\) to \(0\) at stage \(r+1\). This is a concrete reason not to exchange a pointwise limit and an infinite sum unless the hypotheses of a convergence theorem have been checked.

A second useful caution is that “finite partial sum” means a sum over a finite set of indices, not a sum over the first few entries chosen by their size. For nonnegative terms, the definition is a supremum over all finite subsets. An enumeration gives one increasing exhaustion and computes that supremum, but the theorem does not require the terms to be ordered by magnitude. The finite-set formulation also handles the empty index set, finite index sets, zero entries, and infinite entries without changing the statement.