Measurable Functions

Definition and equivalences

Definition (Measurable function)

Let \(f : E \to \overline{\mathbb{R}} = \mathbb{R} \cup \{\pm\infty\}\). We say that \(f\) is measurable if for every \(a \in \mathbb{R}\)

\[\{f > a\} := \{x \in E : f(x) > a\} \in \mathcal{M}.\]

Since \(E = \bigcup_{k=1}^{\infty}\{f > -k\} \cup \{f = -\infty\}\), if \(f\) is measurable then \(E\) is measurable if and only if \(\{f = -\infty\}\) is measurable. In what follows \(E\) is assumed to be measurable.

Example (Continuous and indicator functions)

1. If \(f : \mathbb{R}^{d} \to \mathbb{R}\) is continuous, \(\{f > a\} = f^{-1}((a, \infty))\) is open, hence measurable; every continuous function is measurable.
2. If \(f = \mathbf{1}_{A}\) with \(A\) measurable, then \(\{f > a\}\) equals \(E\) if \(a < 0\), \(A\) if \(0 \leq a < 1\) and \(\emptyset\) if \(a \geq 1\); in every case it is measurable.

Theorem (Equivalent conditions for measurability)

Let \(f : E \to \overline{\mathbb{R}}\) with \(E\) measurable. The following conditions, for every \(a \in \mathbb{R}\), are equivalent:

\[\text{(i) } \{f > a\} \in \mathcal{M}, \quad \text{(ii) } \{f < a\} \in \mathcal{M}, \quad \text{(iii) } \{f \leq a\} \in \mathcal{M}, \quad \text{(iv) } \{f \geq a\} \in \mathcal{M}.\]

Proof: \(\{f \leq a\} = \{f > a\}^{c}\) and \(\{f < a\} = \{f \geq a\}^{c}\), so that (i)\(\Leftrightarrow\)(iii) and (ii)\(\Leftrightarrow\)(iv) by closure under complements. Moreover

\[\{f \geq a\} = \bigcap_{n=1}^{\infty}\Big\{f > a - \tfrac{1}{n}\Big\}, \qquad \{f > a\} = \bigcup_{n=1}^{\infty}\Big\{f \geq a + \tfrac{1}{n}\Big\},\]

which shows (i)\(\Rightarrow\)(iv) and (iv)\(\Rightarrow\)(i) by closure under countable intersections and unions. The four conditions are therefore equivalent.

Note (Other measurable sets associated with \(f\))

If \(f : E \to \overline{\mathbb{R}}\) is measurable, then the following are measurable

\[\{f > -\infty\} = \bigcup_{k}\{f > -k\}, \quad \{f < \infty\} = \bigcup_{k}\{f \leq k\}, \quad \{f = \infty\}, \quad \{a \leq f \leq b\}, \quad \{a \leq f < b\},\]

since they are countable intersections and unions of the sets \(\{f > a\}\), \(\{f \leq a\}\), etc.

Definition (Borel measurable function)

\(f : E \to \mathbb{R}\) is Borel measurable if \(E \in \mathcal{B}\) and \(\{f > a\} \in \mathcal{B}\) for every \(a \in \mathbb{R}\).

Theorem (Measurability via preimages of open sets)

Let \(f : E \to \mathbb{R}\). Then \(f\) is measurable if and only if \(f^{-1}(G)\) is measurable for every open \(G \subseteq \mathbb{R}\).

Proof: If the preimages of open sets are measurable, taking \(G = (a, \infty)\) gives \(\{f > a\} = f^{-1}(G)\) measurable, hence \(f\) is measurable. Conversely, let \(f\) be measurable and \(G \subseteq \mathbb{R}\) open; write \(G = \bigcup_{k}(a_{k}, b_{k})\). Since

\[f^{-1}\big((a_{k}, b_{k})\big) = \{a_{k} < f\} \cap \{f < b_{k}\}\]

is measurable, \(f^{-1}(G) = \bigcup_{k} f^{-1}\big((a_{k}, b_{k})\big)\) is a countable union of measurable sets, hence measurable.

Composition, equality almost everywhere and operations

Lemma (Composition of a measurable function with a continuous one)

Let \(f : E \to \mathbb{R}\) be measurable and \(\phi : \mathbb{R} \to \mathbb{R}\) continuous. Then \(\phi \circ f\) is measurable.

Proof: Let \(G \subseteq \mathbb{R}\) be open. Then \((\phi \circ f)^{-1}(G) = f^{-1}\big(\phi^{-1}(G)\big)\). Since \(\phi\) is continuous, \(\phi^{-1}(G)\) is open, and since \(f\) is measurable, \(f^{-1}\big(\phi^{-1}(G)\big)\) is measurable. By the previous theorem, \(\phi \circ f\) is measurable.

Note (Immediate consequences)

If \(f\) is measurable, then $$

f

\(,\)

f

^{p}\((\)p > 0\(),\)e^{cf}\((\)c \in \mathbb{R}$$) and the positive and negative parts

\[f^{+} = \max\{f, 0\}, \qquad f^{-} = \max\{-f, 0\}\]

are measurable. (Here \(f^{+}, f^{-} \geq 0\), \(f = f^{+} - f^{-}\) and $$

f

= f^{+} + f^{-}$$, in agreement with the notation for the positive and negative parts of a real number.)

Definition (Almost everywhere)

A property holds almost everywhere (a.e.) if it holds except on a set of measure zero.

Lemma (Equality almost everywhere preserves measurability)

Let \(f, g : E \to \mathbb{R}\) with \(f\) measurable and \(g = f\) almost everywhere. Then \(g\) is measurable.

Proof: Let \(a \in \mathbb{R}\). Decomposing according to where \(f\) and \(g\) agree,

\[\{g > a\} = \big(\{g > a\} \cap \{f = g\}\big) \cup \big(\{g > a\} \cap \{f \neq g\}\big).\]

The first set is \(\{f > a\} \cap \{f = g\}\), which is measurable (an intersection of measurable sets, apart from the set \(\{f \neq g\}\) of measure zero, which is measurable). The second is contained in \(\{f \neq g\}\), of measure zero, hence measurable. Therefore \(\{g > a\}\) is a union of measurable sets and \(g\) is measurable.

Lemma (Composition with a continuous function for functions finite a.e.)

Let \(f : E \to \overline{\mathbb{R}}\) be measurable with \(m(\{f = \infty\}) = m(\{f = -\infty\}) = 0\). Then \(\phi \circ f\) is measurable for every continuous \(\phi : \mathbb{R} \to \mathbb{R}\).

Proof: Let \(F = \{x \in E : f(x) \in \mathbb{R}\}\) and define \(f_{1} = f\) on \(F\) and \(f_{1} = 0\) on \(E \setminus F\). Since \(E \setminus F = \{f = \infty\} \cup \{f = -\infty\}\) has measure zero, \(f_{1}\) is measurable and \(f_{1} = f\) almost everywhere. By the composition lemma, \(\phi \circ f_{1}\) is measurable, and since \(\phi \circ f_{1} = \phi \circ f\) a.e., the previous lemma gives that \(\phi \circ f\) is measurable.

Lemma (The set where one measurable function exceeds another is measurable)

Let \(f, g : E \to \mathbb{R}\) be measurable. Then \(\{f > g\}\) is measurable.

Proof: Let \(\{q_{n}\}_{n=1}^{\infty}\) be an enumeration of \(\mathbb{Q}\). Since \(f(x) > g(x)\) if and only if there exists a rational strictly between \(g(x)\) and \(f(x)\),

\[\{f > g\} = \bigcup_{n=1}^{\infty}\big(\{f > q_{n}\} \cap \{q_{n} > g\}\big),\]

a countable union of intersections of measurable sets, hence measurable.

Lemma (Vector space structure of the measurable functions)

Let \(f, g : E \to \mathbb{R}\) be measurable. Then:

1. \(f + \lambda\) and \(\lambda f\) are measurable for every \(\lambda \in \mathbb{R}\);
2. \(f + g\) is measurable.

Proof: Part (i) is left as an exercise (direct from the definition: \(\{f + \lambda > a\} = \{f > a - \lambda\}\) and, for \(\lambda > 0\), \(\{\lambda f > a\} = \{f > a/\lambda\}\), etc.). For (ii), \(g\) and \(\lambda - g\) are measurable by (i), and by the previous lemma applied to \(f\) and \(\lambda - g\),

\[\{f + g > \lambda\} = \{f > \lambda - g\}\]

is measurable for every \(\lambda \in \mathbb{R}\), hence \(f + g\) is measurable.

Exercise (The case of infinite values in the sum)

The previous lemma still holds for \(f, g : E \to \overline{\mathbb{R}}\) provided that \(f + g\) is well defined (that is, avoiding the indeterminate form \(\infty - \infty\)). Prove this variant.

Corollary (Product and quotient of measurable functions)

Let \(f, g : E \to \overline{\mathbb{R}}\) be measurable. Then \(fg\) is measurable (with the convention \(0 \cdot (\pm\infty) = 0\), so that the product is always defined), and \(f/g\) is measurable if \(g \neq 0\).

Proof: Let \(F = \{f \in \mathbb{R}\} \cap \{g \in \mathbb{R}\}\). For \(a \geq 0\),

\[\{fg > a\} = \big(\{fg > a\} \cap F\big) \cup \big(\{f = \infty\} \cap \{g > 0\}\big) \cup \big(\{g = \infty\} \cap \{f > 0\}\big) \cup \big(\{f = -\infty\} \cap \{g < 0\}\big) \cup \big(\{g = -\infty\} \cap \{f < 0\}\big),\]

and every set on the right-hand side is measurable except, a priori, the first. On \(F\) the polarisation identity holds

\[fg = \tfrac{1}{4}\big((f + g)^{2} - (f - g)^{2}\big);\]

since \(f + g\) and \(f - g\) are measurable (previous lemma) and squaring is composition with the continuous map \(t \mapsto t^{2}\), \(fg\) is measurable on \(F\), so that \(\{fg > a\} \cap F\) is measurable. The treatment of \(a < 0\) and of the quotient \(f/g\) (composing with \(t \mapsto 1/t\) where \(g \neq 0\)) is left as an exercise.

Sequences of measurable functions and simple functions

Theorem (Sup, inf, limit superior and limit inferior of measurable functions)

Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of measurable functions \(E \to \overline{\mathbb{R}}\). Then \(\sup_{k} f_{k}\), \(\inf_{k} f_{k}\), \(\limsup_{k} f_{k}\) and \(\liminf_{k} f_{k}\) are measurable.

Proof: For every \(a \in \mathbb{R}\),

\[\Big\{\sup_{k \geq 1} f_{k} > a\Big\} = \bigcup_{k=1}^{\infty}\{f_{k} > a\}\]

is measurable, hence \(\sup_{k} f_{k}\) is measurable. The remaining ones reduce to this one:

\[\inf_{k} f_{k} = -\sup_{k}(-f_{k}), \qquad \limsup_{k} f_{k} = \inf_{k \geq 1}\sup_{\ell \geq k} f_{\ell}, \qquad \liminf_{k} f_{k} = \sup_{k \geq 1}\inf_{\ell \geq k} f_{\ell}.\]

Definition (Simple function)

A function \(\phi\) is simple if there exist measurable sets \(A_{1}, \dots, A_{m}\) and reals \(a_{1}, \dots, a_{m}\) with

\[\phi = \sum_{k=1}^{m} a_{k}\, \mathbf{1}_{A_{k}}.\]

Exercise (Canonical form of a simple function)

Prove that every simple function \(\phi\) admits a representation \(\phi = \sum_{k=1}^{\ell} b_{k}\, \mathbf{1}_{B_{k}}\) with \(B_{1}, \dots, B_{\ell}\) measurable and pairwise disjoint and \(b_{1}, \dots, b_{\ell}\) pairwise distinct.

Note (Dyadic approximation of a non-negative function)

Let \(f : E \to [0, \infty]\). For \(k \in \mathbb{N}\) define

\[f_{k} = k\,\mathbf{1}_{B_{k}} + \sum_{j=1}^{k 2^{k}} \frac{j-1}{2^{k}}\,\mathbf{1}_{A_{k}^{j}}, \qquad B_{k} = f^{-1}\big([k, \infty]\big), \quad A_{k}^{j} = f^{-1}\Big(\Big[\tfrac{j-1}{2^{k}}, \tfrac{j}{2^{k}}\Big)\Big),\]

that is, \(f_{k}(x) = \tfrac{j-1}{2^{k}}\) if \(\tfrac{j-1}{2^{k}} \leq f(x) < \tfrac{j}{2^{k}}\) and \(f_{k}(x) = k\) if \(f(x) \geq k\). If \(0 \leq f(x) < k\), then \(0 \leq f(x) - f_{k}(x) < \tfrac{1}{2^{k}}\), so that \(f_{k}(x) \to f(x)\) when \(f(x) < \infty\); and if \(f(x) = \infty\), \(f_{k}(x) = k \to \infty\). Moreover \(f_{k} \leq f_{k+1}\): on passing from \(k\) to \(k+1\) each dyadic interval is split in two, and the value of \(f_{k+1}\) is equal to or one dyadic unit greater than that of \(f_{k}\). If \(f\) is measurable, the \(A_{k}^{j}\) and \(B_{k}\) are measurable and each \(f_{k}\) is simple and measurable.

Theorem (Approximation by simple functions)

Let \(f : E \to \overline{\mathbb{R}}\). Then there exists a sequence of simple functions \(\psi_{k}\) with \(\psi_{k} \to f\) pointwise. If \(f \geq 0\), the sequence may be taken increasing; if \(f\) is measurable, the \(\psi_{k}\) may be taken measurable.

Proof: For \(f \geq 0\) the dyadic sequence \(\psi_{k} = f_{k}\) of the previous note is simple, increasing and converges pointwise to \(f\) (and measurable if \(f\) is). For \(f\) of arbitrary sign one applies the above to \(f^{+}\) and \(f^{-}\) and takes \(\psi_{k} = (f^{+})_{k} - (f^{-})_{k}\); the details of this case are left as an exercise.