A question on an equivalent formulation of bounded unsigned measurable function

Question

The problem is:

Let $f: \mathbb{R}^d \to [0,+\infty]$. Show that the following two statements are equivalent.

$(i)$ $f$ is a bounded unsigned measurable function.

$(ii)$ $f$ is the uniform limit of a sequence of bounded unsigned simple functions.

I've done the $(ii) \to (i)$ part which was straightforward, but have confusion on the $(i) \to (ii)$ part. We're given that $f$ is a bounded unsigned measurable function. Since it is an unsigned measurable function, we can write it as $$f=\lim_{n \to \infty}f_n$$ where for all $n$, $f_n$ is an unsigned simple function. I've taken care of the bounded part in the following way (please see if there's any technical flaw here):

For every $\epsilon \in \mathbb{R}$ and $x \in \mathbb{R}^d$, $\exists~ N=N(\epsilon, x) \in \mathbb{N}$, which may depend on $x$, such that $$\lvert f_n(x)-f(x)\rvert \leq \epsilon ~\forall~ n \geq N \cdots (\star)$$ We know that $f$ is bounded. So $\exists~ B \in \mathbb{R}$ such that $$\lvert f(x) \rvert \leq B ~\forall~ x \in \mathbb{R}^d$$ Coupling this with $(\star)$ and employing the triangular inequality it is easy to see that $f_n$ is bounded, for all $n \in \mathbb{N}$.

The only thing left is to show that the convergence is uniform. What can I do here? We need to get rid of the dependence of $N$ in $(\star)$ on the choice of $x$. My guess is that we can use the boundedness of the functions to show that $\sup_{x \in \mathbb{R}^d} N(\epsilon, x)=N(\epsilon)$ is a finite quantity and we can use it in $(\star)$ to show uniform convergence. But how to do it in a rigorous way? Maybe this line of attack is trivial or maybe it is completely out-of-track, so apology in advance. Any help would be greatly appreciated.

Answer 1

There is a construction that shows this direction:

1. $f$ is bounded, wlog say its bounded by 1. For every natural $n$ break the interval $[0,1]$ into $2^n$ intervals of length $2^{-n}$.
2. Label $A_i$ the $f$-preimage of $[i2^{-n},(i+1)2^{-n}]$, $i=0,\dots 2^n$.
3. Now you define a simple function $\phi _n$ as $$\phi _n=\sum _{i=1} ^{2^n} 2^{-n}i\chi _{A_i}$$
4. The $\phi _n$ converge uniformly to $f$