last updated: May 9, 2024

Probability with Measure

3.5 Simple functions

In this section we are interested in the following class of functions.

  • Definition 3.5.1 Let \((S,\Sigma )\) be a measurable space. We say that a function \(f:S\to \R \) is simple if it has the form

    \begin{equation} \label {eq:simple_def} f = \sum _{i=1}^{n}c_{i}{\1}_{A_{i}} \end{equation}

    where \(c_i\in \R \) and \(A_i\in \Sigma \), with \(A_{i} \cap A_{j} = \emptyset \) whenever \(i \neq j\).

In words, a simple function is a (finite) linear combination of indicator functions of non-overlapping measurable sets. It follows from Example 3.1.2 and Theorem 3.1.5 that every simple function is measurable. Exercise 3.5 shows that sums and scalar multiples of simple functions are themselves simple, so the set of all simple functions forms a vector space.

Our next theorem explains the purpose of simple functions. They are a natural class of functions with which to approximate measurable functions. This will be a key ingredient of Lebesgue integration. Loosely, we will (in Chapter 4) specify how to integrate simple functions, and then use an approximation scheme to extend the same idea to measurable functions.

Our pointwise notation for functions, from Section 2.3, is also useful for inequalities involving functions. For example, we say that \(f:S \rightarrow \R \) is non-negative if \(f\geq 0\), meaning that \(f(x) \geq 0\) for all \(x \in S\). Similarly, we write \(f \leq g\) to mean that \(f(x)\leq g(x)\) for all \(x\). It is easy to check that a simple function of the form (3.4) is non-negative if and only if \(c_{i} \geq 0\) for all \(i\).

  • Theorem 3.5.2 Let \(f:S \rightarrow \R \) be measurable and non-negative. Then there exists a sequence \((s_{n})\) of non-negative simple functions on \(S\) with \(s_{n} \leq s_{n+1} \leq f\) for all \(\nN \) so that \((s_{n})\) converges pointwise to \(f\) as \(n \rightarrow \infty \). Moreover, if \(f\) is bounded then the convergence is uniform.

Proof: We split this into three parts.

Step 1: Construction of \((s_{n})\). Divide the interval \([0, n)\) into \(n2^{n}\) subintervals \(\{I_{j}, 1 \leq j \leq n2^{n}\}\), each of length \(\frac {1}{2^{n}}\) by taking \(I_{j} = \left [\frac {j-1}{2^{n}}, \frac {j}{2^{n}}\right ).\) Let \(E_{j} = f^{-1}(I_{j})\) and \(F_{n} = f^{-1}([n, \infty ))\). Then \(S = \bigcup _{j=1}^{n2^{n}}E_{j} \cup F_{n}\). We define for all \(x \in S\)

\[ s_{n}(x) = \sum _{j=1}^{n2^{n}}\left (\frac {j-1}{2^{n}}\right ){\1}_{E_{j}}(x) + n{\1}_{F_{n}}(x).\]

Step 2: Properties of \((s_{n})\). For \(x \in E_{j}, s_{n}(x) = \frac {j-1}{2^{n}}\) and \(\frac {j-1}{2^{n}} \leq f(x) < \frac {j}{2^{n}}\) and so \(s_{n}(x) \leq f(x)\). For \(x \in F_{n}, s_{n}(x) = n\) and \(f(x) \geq n\). So we conclude that \(s_{n} \leq f\) for all \(\nN \).

To show that \(s_{n} \leq s_{n+1}\), fix an arbitrary \(j\) and consider \(I_{j} = \left [\frac {j-1}{2^{n}}, \frac {j}{2^{n}}\right )\). For convenience, we write \(I_{j}\) as \(I\) and we observe that \(I = I_{1}\cup I_{2}\) where \(I_{1} = \left [\frac {2j-2}{2^{n+1}}, \frac {2j-1}{2^{n+1}}\right )\) and \(I_{2} = \left [\frac {2j-1}{2^{n+1}}, \frac {2j}{2^{n+1}}\right )\). Let \(E = f^{-1}(I), E_{1} = f^{-1}(I_{1})\) and \(E_{2} = f^{-1}(I_{2})\). Then \(s_{n}(x) = \frac {j-1}{2^{n}}\) for all \(x \in E, s_{n+1}(x) = \frac {j-1}{2^{n}}\) for all \(x \in E_{1}\), and \(s_{n+1}(x) = \frac {2j-1}{2^{n+1}}\) for all \(x \in E_{2}\). It follows that \(s_{n} \leq s_{n+1}\) for all \(x \in E\). A similar (easier) argument can be used on \(F_{n}\).

Step 3: Convergence of \((s_{n})\). Fix any \(x \in S\). Since \(f(x) \in \R \) there exists \(n_{0} \in \mathbb {N}\) so that \(f(x) \leq n_{0}\). Then for each \(n > n_{0}, f(x) \in I_{j}\) for some \(1 \leq j \leq n2^{n}\), i.e. \(\frac {j-1}{2^{n}} \leq f(x) < \frac {j}{2^{n}}\). But \(s_{n}(x) = \frac {j-1}{2^{n}}\) and so \(|f(x) - s_{n}(x)| < \frac {1}{2^{n}}\) and pointwise convergence follows. Note that if \(f\) is bounded we can find \(n_{0} \in \mathbb {N}\) so that \(f(x) \leq n_{0}\) for all \(x \in \R \). Then the argument just given yields \(|f(x) - s_{n}(x)| < \frac {1}{2^{n}}\) for all \(x \in \R \) which gives uniform convergence   ∎