Measurable functions §

Equivalent characterizations for $f: X \to \mathbb{R}$:

$f$ is measurable if and only if any (hence all) of the following hold:

1. $\{x : f(x) > a\} \in \mathcal{F}$ for all $a \in \mathbb{R}$
2. $\{x : f(x) \geq a\} \in \mathcal{F}$ for all $a \in \mathbb{R}$
3. $\{x : a \leq f(x) < b\} \in \mathcal{F}$ for all $a, b \in \mathbb{R}$

Key facts:

• If $f$ and $g$ are measurable, then $f + g$, $fg$, $|f|$, $f^+ = \max(f,0)$, $f^- = \max(-f,0)$ are all measurable.
• See Pointwise limits for closure under limits, and Simple functions for the approximation theorem.

Note on the nature of measurability. Measurability is a relationship between two measurable structures (the sigma-algebra on the domain and the sigma-algebra on the codomain), not a property of the function in isolation. The same function $f: X \to \mathbb{R}$ may be measurable with respect to one sigma-algebra on $X$ and not another.

Non-measurable functions. If $\mathcal{F} = \{\varnothing, X\}$ (the trivial sigma-algebra), then the only measurable functions $f: X \to \mathbb{R}$ are the constant functions: for any non-constant $f$, there exists a Borel set $B$ such that $f^{-1}(B)$ is a proper non-empty subset of $X$, which is not in $\mathcal{F}$. More dramatically, on $(\mathbb{R}, \mathcal{L})$ (Lebesgue-measurable sets), the indicator of a Vitali set $V$ is not measurable because $V \notin \mathcal{L}$.

Simple functions are the building blocks of the Lebesgue integral. Every nonnegative measurable function $f$ is the pointwise limit of an increasing sequence of simple functions $f_n \uparrow f$.

Example. On $\mathbb{R}$, the floor function $f(x) = \lfloor x \rfloor$ is not simple (it takes infinitely many values). However, restricted to $[0, 3)$ it is simple: $f = 0 \cdot \mathbf{1}_{[0,1)} + 1 \cdot \mathbf{1}_{[1,2)} + 2 \cdot \mathbf{1}_{[2,3)}$

The integral of a simple function is defined algebraically (see Lebesgue integral for the full construction): $\int f \, d\mu = \sum_{i=1}^{n} a_i \, \mu(A_i)$

This is independent of the particular representation of $f$ as a sum of indicator functions.

Proof. For any $C \in \mathcal{H}$ (the sigma-algebra on $Z$): $(g \circ f)^{-1}(C) = f^{-1}(g^{-1}(C))$ Since $g$ is measurable, $g^{-1}(C) \in \mathcal{G}$. Since $f$ is measurable, $f^{-1}(g^{-1}(C)) \in \mathcal{F}$.

Why the corollary matters. Every continuous function $g: \mathbb{R} \to \mathbb{R}$ is Borel-measurable (preimages of open sets are open, hence Borel). So if $f$ is measurable, then $\sin(f)$, $e^f$, $|f|$, $f^2$, etc. are all measurable. This is the practical workhorse: you rarely need to check measurability from scratch; you compose known measurable functions.

Proof sketch. For $\sup_n f_n$: $\{x : \sup_n f_n(x) > a\} = \bigcup_n \{x : f_n(x) > a\}$, which is a countable union of measurable sets.

This theorem is crucial because it means the class of measurable functions is “closed under the operations we actually use”: taking limits, composing, adding, multiplying.

Connection to the Riemann integral. Every Riemann-integrable function is Lebesgue-measurable (it is the a.e. limit of step functions, which are simple functions). The converse is false: the Dirichlet function $\mathbf{1}_{\mathbb{Q}}$ is measurable (hence Lebesgue-integrable, with integral 0) but not Riemann-integrable.

Step 1: Show measurability.

$f$ takes only the values 0 and 1. For any Borel set $B \subseteq \mathbb{R}$:

• If $0 \in B$ and $1 \in B$: $f^{-1}(B) = \mathbb{R} \in \mathcal{B}(\mathbb{R})$ ✓
• If $0 \in B$, $1 \notin B$: $f^{-1}(B) = \mathbb{R} \setminus \mathbb{Q} \in \mathcal{B}(\mathbb{R})$ (complement of a countable set) ✓
• If $0 \notin B$, $1 \in B$: $f^{-1}(B) = \mathbb{Q} \in \mathcal{B}(\mathbb{R})$ (countable set) ✓
• If $0 \notin B$, $1 \notin B$: $f^{-1}(B) = \varnothing \in \mathcal{B}(\mathbb{R})$ ✓

So $f$ is Borel-measurable.

Step 2: Construct simple functions.

Let $f_n(x) = \mathbf{1}_{\{q_1, \ldots, q_n\}}(x)$ where $\{q_1, q_2, \ldots\}$ is an enumeration of $\mathbb{Q}$. Each $f_n$ is a simple function (it takes values 0 and 1 on the measurable sets $\{q_1, \ldots, q_n\}$ and its complement). And $f_n(x) \to \mathbf{1}_{\mathbb{Q}}(x) = f(x)$ for every $x$.