# Measurability of a function

Let $f:X \to \mathbb{R}^m$ be Borel-measurable and $z_h \in S^{m-1} \forall h \in \mathbb{N}$ (Here $S^{m-1}$ is the unit sphere in $\mathbb{R}^m$). Fix an $\epsilon>0$ and define : $\sigma : X \to \mathbb{N}$, $\sigma(x) := \min \{h\in \mathbb{N}: \langle f(x),z_h\rangle \geq (1-\epsilon)|f(x)|\} \forall x \in X$

Is $\sigma$ measurable? (The authors take this is as measurable in the text "Functions of Bounded Variation and Free Discontinuity Problems" by Luigi Ambrosio et al. page 11)

I tried to simplify the problem as follows:- $\sigma(x) := \min \{h\in \mathbb{N}: g_h(x) \geq 0\} \forall x \in X$ for some measurable functions $g_h$. Therefore, $\sigma^{-1}(h) = (g_{h_1}^{-1}[0,\infty) \cap g_{h_2}^{-1}[0,\infty) ... \cap g_{h_n}^{-1}[0,\infty)... )\cap (g_{h-1}^{-1}[0,\infty)^c \cap g_{h-2}^{-1}[0,\infty)^c ...\cap g_{0}^{-1}[0,\infty)^c)$, where $h_1, h_2, ...$ are all greater than or equal to $h$ and atleast one of them equals $h$. I see that the number of such $\{h_1,h_2,...\}$ is uncountable. So I cant prove that $\sigma$ is measurable this way. Please let me know if I can prove this any other way.

Any help is greatly appreciated.

Thanks, Phanindra

-
Why don't we have $\sigma^{-1}(h)=\bigcap_{j=0}^{h-1}g_j(\left[0,\infty\right)^c)\cap g_h^{-1}(\left[0,\infty\right))$? – Davide Giraudo Aug 29 '11 at 9:11
Or, equivalently, $[\sigma\ge h+1]=\bigcap_{k=0}^h[g_k<0]$. – Did Aug 29 '11 at 9:54
Davide, Didier: Thanks for the answers. They are very clear. I have confused myself with all unnecessary details :). – jpv Aug 29 '11 at 16:37
jpv, thanks. // Please use the @ thing, otherwise your comment will not reach us (I read this by chance). – Did Aug 30 '11 at 15:36

• Since $\sigma$ is $\Bbb N$ valued, it's enough to show that for a fixed integer $n$, the set $\sigma^{-1}(\{n\})$ is measurable (because if $A\subset \Bbb N$, then $\sigma^{-1}(A)=\bigcup_{n\in A}\sigma^{-1}(\{n\}$ will be measurable as a countable union of such sets).
• We can write $\sigma^{-1}(\{n\})=\bigcap_{j=0}^{n-1}g_j([0,+\infty))\cap g_n^{—1}([0,+\infty))$ with $g_j(x):=\langle f(x),z_j\rangle-(1-\varepsilon)|f(x)|$, a measurable map.