In this brief note and the next, we take a bottom-up approach to defining the Lebesgue integral on an arbitrary measure space.

Definition

Let the triple \((X,\mathfrak{M},\mu)\) denote this measure space: \(X\) is the underlying set, \(\mathfrak{M}\) is the \(\sigma\)-algebra of measurable subsets, and \(\mu\) is a (positive) measure on \((X,\mathfrak{M})\). The atoms of integration on \(X\) are the characteristic functions associated to the measurable subsets. In the context of a finite measure with total measure \(\mu(X) = 1\), these integrals admit interpretation as probabilities of events. Declare

\[\int \chi_{E} = \mu(E)\]

for any measurable \(E \in \mathfrak{M}\). Linearity of \(\int\) forces the relation

\[\int \sum_{j = 1}^{m} a_{j}\chi_{E_{j}} = \sum_{j = 1}^{m} a_{j} \mu(E_{j})\]

for any finite collection \(E_{1},\dots,E_{m}\) in \(\mathfrak{M}\) and coefficients \(a_{1},\dots,a_{m}\) in the nonnegative extended real numbers. Such linear combinations are called simple functions. Alternatively, simple functions are those functions to \([0,\infty]\) that assume but finitely many values. N.B., We adopt the convention \(\infty \cdot 0 = 0\).

Here, we must contend with the fact that the canonical map from the free \(\mathbb{R}\)-vector space generated by \(\mathfrak{M}\) to the \(\mathbb{R}\)-vector space of functions on \(X\) is not monic; the kernel is generated by pre-existing algebraic structure on \(\mathfrak{M}\).

For instance, we can take \(X\) to be the outcome space of a flip of a fair coin.

\[X = \{H,T\}.\]

The constant function \(1\), i.e., the indicator of the sure event, admits the following two representations.

\[1 = \chi_{X} = \chi_{\{H\}} + \chi_{\{T\}}.\]

Note the equivalence of these two representations ultimately arises from the relation

\[X = \{H\} \cup \{T\}\]

on \(\mathfrak{M}\). We thereby witness that it is the additivity of the measure \(\mu\) that ensures \(\int\) is well-defined.

Proof

Consider

\[f = \sum_{j = 1}^{m} a_{j}\chi_{E_{j}}.\]

We reduce to the case when the \(E_{j}\) partition the space \(X\) as follows. For \(\epsilon \in \{0,1\}^{m}\), declare

\[E(\epsilon) = \bigcap_{j = 1}^{m} \begin{cases} E_{j} & \text{if $\epsilon_{j} = 1$,} \\ E_{j}^{c} & \text{if $\epsilon_{j} = 0$.} \end{cases}\]

The \(E(\epsilon)\) partition \(X\). Further, on each \(E(\epsilon)\), the function \(f\) is constant and assumes the value \(\sum_{j = 1}^{m} a_{j}\epsilon_{j}\). Hence,

\[\sum_{j = 1}^{m} a_{j}\chi_{E_{j}} = \sum_{\epsilon \in \{0,1\}^{n}} \left( \sum_{j = 1}^{m} a_{j}\epsilon_{j} \right) \chi_{E(\epsilon)}.\]

Likewise,

\[\sum_{j = 1}^{m} a_{j} \mu(E_{j}) = \sum_{j = 1}^{m} a_{j} \sum_{\epsilon \in \{0,1\}^{m}, \epsilon_{j} = 1} \mu(E(\epsilon)) = \sum_{\epsilon \in \{0,1\}^{m}} \left( \sum_{j = 1}^{m} a_{j}\epsilon_{j} \right) \mu(E(\epsilon)).\]

In sum, it suffices to show that any representation of \(f\) by a partition of \(X\) yields the same integral.

Assume

\[f = \sum_{j = 1}^{m} a_{j}\chi_{E_{j}} = \sum_{k = 1}^{n} b_{j}\chi_{F_{j}},\]

where \((E_{j})_{j = 1}^{m}\) and \((F_{k})_{k = 1}^{n}\) are both partitions of \(X\).

\[\sum_{j = 1}^{m} a_{j}\mu(E_{j}) = \sum_{j = 1}^{m} a_{j} \sum_{k = 1}^{n} \mu(E_{j} \cap F_{k})\] \[= \sum_{j = 1}^{m} \sum_{k = 1}^{n} a_{j} \mu(E_{j} \cap F_{k})\] \[= \sum_{j = 1}^{m} \sum_{k = 1}^{n} b_{k} \mu(E_{j} \cap F_{k})\] \[= \sum_{k = 1}^{n} b_{k} \sum_{j = 1}^{m} \mu(E_{j} \cap F_{k})\] \[= \sum_{k = 1}^{n} b_{k} \mu(F_{k}).\]

That is, the value of the integral is independent of the choice of partition. \(\tag*{$\blacksquare$}\)

Remarks

  • The main idea underpinning the argument is the fact that the measure of a set can be computed via any partition.
  • Folland defines the integral of a simple function in terms of its standard representation, i.e., \(f = \sum_{a \in \mathbb{R}} a \chi_{f^{-1}(a)}.\) The preceding argument demonstrates that, in fact, any representation of a simple function by finitely many characteristic functions will yield the same result.
  • The result is equivalent to the additivity (2.13b) of Folland’s integral.