Measurablity of inner product If $f : X \to \mathbb{R}^n$ is measurable, then $\langle f,f\rangle = ||f||_2^2: X \to \mathbb{R}$ is measurable (if $\langle f,f\rangle < c$ for $c > 0$ then $f$ should lie in an open ball with the origin as center and radius $\sqrt{c}$. The pre-image of which is measurable).
What can be said about $\langle f,z\rangle$ where $z \in \mathbb{R}^n$ is an arbitrary vector? Is it still measurable?
(A general query: I cannot access this website from my work place, where can I get help?)
 A: There seems to be some confusion here, so let me answer instead of commenting:
If $f: (X,\Sigma_{X}) \to (Y,\Sigma_{Y})$ is a function between measurable spaces, it is called measurable if $f^{-1}(S) \in \Sigma_{X}$ for all $S \in \Sigma_{Y}$. So much for the general definition.
Now if $f: (X, \Sigma_{X}) \to Y$ is a map to some (say separable metric) space then measurability always means measurable with respect to the Borel $\sigma$-algebra $\mathcal{B}_{Y}$ on $Y$ unless otherwise specified.
The reason for this is simple: we want at least continuous maps to be measurable. Even for a continuous $f: [0,1] \to [0,1]$, the pre-image of a Lebesgue measurable set need not be Lebesgue measurable, and that should probably be reason enough for dismissing the notion of Lebesgue-Lebesgue measurable maps as hopelessly useless (at least from this perspective).
It is an interesting (but somewhat non-trivial) exercise to determine all maps $f: [0,1] \to [0,1]$ with the property that $f^{-1}(L)$ is Lebesgue measurable for all Lebesgue measurable sets $L$. I don't want to give too much away, but a small hint anyway: By a theorem of Vitali a subset $S$ of $[0,1]$ has the property that all its subsets are Lebesgue measurable if and only if $S$ has measure zero.

We can't compose Lebesgue measurable functions without thinking hard:
if $f:[0,1] \to [0,1]$ is continuous and $g: [0,1] \to [0,1]$ is Lebesgue measurable in the sense that $L = g^{-1}(B)$ is Lebesgue measurable for all Borel sets $B$ then we're doomed: We simply can't say anything about $f^{-1}(L) = f^{-1}(g^{-1}(B))$ without knowing more about $f$ and $g$.
Fortunately, we're in the other situation: we're given that $f: (X,\Sigma_{X}) \to \mathbb{R}^{n}$ is measurable, so $f^{-1}(B) \in \Sigma_{X}$ for all Borel sets. We also have a Borel measurable (even continuous) map $g: \mathbb{R}^{n} \to \mathbb{R}$ and everything works out the way we want it to be: $g^{-1}(B)$ is Borel, hence $f^{-1}(g^{-1}(B)) \in \Sigma_X$, as we wanted. 
