Measurable function

A function $f:(X,\mathcal{A})\to (Y,\mathcal{B})$ between measurable spaces is measurable when $f^{-1}(B)\in \mathcal{A}$ for every $B\in \mathcal{B}$.

A Lebesgue measurable function $f:\mathbb{R}\to \mathbb{R}$ has a Lebesgue measurable set as the preimage of every open subset of $\mathbb{R}$. Equivalently, use every Borel set in the target: the sets whose preimages are measurable form a sigma-algebra containing all open sets. The source carries the Lebesgue sigma-algebra and the target its Borel sigma-algebra. A Borel measurable function on $\mathbb{R}$ instead requires the preimages to be Borel in the source as well. Restricting the source to a Lebesgue measurable set uses its induced sigma-algebra.

For real random variables $X,Z$, if $Z$ is measurable with respect to the sigma-algebra generated by a random variable $X$, there is a Borel measurable function $g:\mathbb{R}\to \mathbb{R}$ with $Z=g(X)$. To prove the bounded case, approximate $Z$ pointwise by finite-valued simple functions $Z_m$. Each level set is $X^{-1}(B)$ for a Borel set $B$, so each approximation factors as $Z_m=g_m(X)$ with $g_m$ a Borel measurable function. Apply scalar clipping to $g_m$ with a common bounded interval and use its limit superior to obtain $g$. For unbounded finite $Z$, apply the bounded case to the one-to-one bounded continuous function $t\mapsto \arctan t$, then invert on $(-\pi /2,\pi /2)$ and assign arbitrary values elsewhere. Exact measurability gives exact factorization; assumptions after adjoining null sets give an almost surely equal version.

The essential range of a real or complex measurable function $m$ consists of the values $z$ whose every neighborhood receives inputs of positive measure:It depends on the measure as well as the function. This definition concerns values arbitrarily near $z$, not necessarily inputs at which $m=z$ exactly. It therefore retains meaningful limiting values while ignoring values occurring only on null sets. ^[ref].

For $m(x)=x$ on $(0,1)$ with Lebesgue measure, the ordinary image of a function is $(0,1)$, but the essential range is $[0,1]$. Neither endpoint is attained, yet inputs with $|m(x)|<\epsilon$ or $|m(x)-1|<\epsilon$ have positive measure for every $\epsilon >0$. Every interior value also belongs to the essential range, although its exact fiber is a singleton of zero measure. Positive measure of a fiber is sufficient, but is not necessary, for a value to belong to the essential range.

The two-sided example $g$ from essential infimum and essential supremum illustrates what disappears. It follows $g(x)=x$ on $[0,1]$, except $g(1/4)=-3$ and $g(3/4)=4$. Its ordinary image of a function is $([0,1]\setminus \{1/4,3/4\})\cup \{-3,4\}$, whereas its essential range remains $[0,1]$. The missing ordinary values $1/4$ and $3/4$ still have arbitrarily close neighbors with positive measure; the exceptional values $-3$ and $4$ can each be isolated by a sufficiently small neighborhood whose preimage is a null set.

For the three-point example in L-infinity norm, the values are $1,3,10$. With weights $(0.5,0.5,0)$ the essential range is $\{1,3\}$; with weights $(0.499,0.5,0.001)$ it is $\{1,3,10\}$. A small but positive probability retains ten, so an essential range does not discard a value merely because it is rare. For a real function on a space of positive total measure, the essential infimum and essential supremum are the lower and upper edges of its essential range, allowing infinite bounds. The L-infinity norm instead takes the upper edge of the essential range of $|m|$.

A useful operator connection is the spectrum of a real multiplication operator: on a nonzero L2 space over a sigma-finite measure, multiplication by a real measurable function $m$ has spectrum equal to its essential range, with the natural domain when $m$ is unbounded. Outside that range, $|m-z|$ has a positive lower bound almost everywhere, so multiplication by $1/(m-z)$ gives an inverse function that is a bounded linear operator. Near an essential value, this lower bound fails. Thus discarding null sets is precisely what makes the range match an operator acting on almost everywhere equivalence classes. ^[ref].

Measurability is the property that inverse images of measurable target sets are measurable in the source space. For a stochastic process, joint measurability in time and sample is a different condition from the separate adaptedness condition at each fixed time. The predictable sigma-algebra and progressive measurability encode particular joint measurability requirements.

For topological spaces $X,Y$, a Borel measurable function $f:X\to Y$ is a measurable function for the Borel sigma-algebras on both spaces. Equivalently, the preimage of every open set in $Y$ is a Borel set in $X$. With a different source sigma-algebra $\mathcal{A}$, the phrase measurable into $Y$ with its Borel sigma-algebra means $f^{-1}(B)\in \mathcal{A}$ for every Borel $B\subseteq Y$; specifying the source is essential.

If $f:(X,\mathcal{A})\to (Y,\mathcal{B})$ and $g:(Y,\mathcal{B})\to (Z,\mathcal{C})$ are measurable, then $g\circ f$ is measurable because

If $g$ is a Lebesgue measurable function and $h$ is a Borel measurable function, then $h\circ g$ is a Lebesgue measurable function. For an open set $U$, $h^{-1}(U)$ is a Borel set. By measurability of $g$, $g^{-1}(h^{-1}(U))$ is a Lebesgue measurable set, and it equals $(h\circ g{)}^{-1}(U)$. This is composition of measurable functions with the intermediate Borel sigma-algebra stated explicitly. Merely knowing both functions are Lebesgue measurable would not ensure that the intermediate sigma-algebras match.