Let $(X,\Sigma)$ be a measurable space and $f:X\to \mathbb{R}^d$ be a measurable function. Suppose that $\Vert \cdot \Vert $ is a norm of $\mathbb{R}^d$. How can I show that there's a sequence $(\varphi_n)_{n\in\mathbb{N}}$ of simple functions such that $\lim_{n\to\infty}\varphi_n(x)=f(x)$ and $\Vert \varphi _n(x)\Vert \leq \Vert \varphi _{n+1}(x)\Vert \leq \Vert f(x)\Vert $ for any $x\in X$ and $n\in\mathbb{N}$?
I know how to prove the existence of that sequence if $\Vert \cdot \Vert $ is the Euclidean norm, but I don't know how to prove that for any norm of $\mathbb{R}^d$.
Suppose that $\pi_i :\mathbb{R}^d\to \mathbb{R}$ is the $i$th canonical projection for all $i\in \{1,\cdots,d\}$.
I know that for all $i\in \{1,\cdots, d\}$ there's a sequence of simple functions $(\varphi_n^i)_{n\in\mathbb{N}}$ such that for all $x\in X$ we have $\lim_{n\to\infty}\varphi _n^i(x)=(\pi _i\circ f)(x)$ and $|\varphi _n^i(x)|\leq |\varphi _{n+1}^i(x)|\leq |(\pi _i\circ f)(x)|$ for all $n\in\mathbb{N}$.
I defined $\varphi _n:X\to \mathbb{R}^d$ by $\varphi _n(x):=(\varphi _n^1(x),\cdots, \varphi _n^d(x))$ for all $n\in\mathbb{N}$ and tried to conclude that $\Vert \varphi _n(x)\Vert \leq \Vert \varphi _{n+1}(x)\Vert \leq \Vert f(x)\Vert $ for any $x\in X$ and $n\in\mathbb{N}$, however I failed.
I also tried to use the fact the every norm of $\mathbb{R}^d$ is equivalent to the Euclidean norm with the fact that $a\varphi $ is simple for any $a\in\mathbb{R}$ and simple function $\varphi $, but I always end up with a annoying constant in that inequality.
Thank you for your attention!
Take $n$, split $[-n,n]^d$ into $(2n^2)^d$ cubes with side-length $1/n$. If $C$ is such a cube then let $x_C$ be an element of least norm in the cube.
Now take $x\in X$. If $f(x)\in C$ for some of these cubes set $\phi_n(x):=x_C$. If $f(x)\not\in [-n,n]^d$ then $\phi_n(x):=x_C$, where $C$ is a cube that contains a point of $[-n,n]^d$ that is closest to $f(x)$.
If $n > \|f(x)\|$ then $|f(x) - \phi_n(x)| \le diam( [0,1/n]^d)$, which tends to zero for $n\to\infty$. Also $\|\phi_n(x)\| \le \|f(x)\|$.
To get monotonicity, we restrict the sequence to powers of $2$. Set $\psi_k := \phi_{2^k}$. Then the cubes to side-length $1/2^{k+1}$ are contained in cubes of side-length $1/2^k$.
Take $x\in X$, $f(x) \in C$ with $C$ a cube of side-length $1/2^{k+1}$, $f(x) \in C'$ with $C'$ a cube of side-length $1/2^k$. Then by construction $C' \supset C$ and $\|x_C\| \ge \|x_{C'}\|$, $\psi_k(x) = x_{C'}$, $\psi_{k+1}(x)=x_C$, and $\|\psi_k(x)\|\le \|\psi_{k+1}(x)\|$. And $(\psi_k)$ has the desired properties.
This mimics more or less the standard proof for the $d=1$ case. Only difference is that we do not explicitly calculate the elements of least norm in the cube.