Exercise 1.3.4 (Bounded unsigned Lebesgue measurable functions)

• Suppose that ${(f_n)}_{n\in \mathbb{N}}$ is a sequence of unsigned bounded functions which converge to $f$:

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->𝜖>0\exists <!--\; FUNCTION\; APPLICATION\; -->N\in \mathbb{N}\forall <!--\; FUNCTION\; APPLICATION\; -->n\ge N:d_{\infty }(f_n,f)\le 𝜖$$

  and we also assumed that $d_{\infty }(f_n,0)\le M_n\in ℝ$ for all $n\in \mathbb{N}$. By the triangle inequality we thus have

  $$d_{\infty }(f,0)\le d_{\infty }(f,f_n)+d_{\infty }(f_n,0)$$

  and since both quantities are finite, we conclude that $f$ is bounded by as well. Since uniform convergence implies pointwise convergence, $f$ must be unsigned Lebesgue measurable.

• By Lemma 1.3.9 (iv) $f$ is the supremum of an increasing sequence ${(f_n)}_{n\in \mathbb{N}}$ of (1) unsigned simple (2) bounded functions with (3) finite measure support. This has already proven most of the properties of ${(f_n)}_{n\in \mathbb{N}}$ that we needed to demonstrate. The only thing that is left to verify is the uniform convergence. The only additional assumption that we have here and did not have in Lemma 1.3.9 (iv) is that $f$ is bounded; obviously we need to profit from this. Pick an arbitrary $𝜖>0$. By the finiteness of $f$ we have for all $x\in {ℝ}^d$:

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->𝜖>0\exists <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N}:d(f_n(x),f(x))\le 𝜖$$

  But we only have finite number of values $c_1,\dots ,c_m$ that $f_n$ can take on, and we can divide the domain ${ℝ}^d$ into $f^{-1}(c_1)\cup \dots \cup f^{-1}(c_n)$. By the pointwise convergence (supremum) we can find $m$ values $n_1,\dots ,n_m$ such that $d(f_n(x),f(x))\le 𝜖$ for all $x\in f^{-1}(c_i)$. Set $N:=\max <!--\; FUNCTION\; APPLICATION\; -->\{n_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,n_m\}$. We then have

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->𝜖>0\exists <!--\; FUNCTION\; APPLICATION\; -->N\in \mathbb{N}:d(f_n(x),f(x))\le 𝜖\text{ for all }x\in {ℝ}^d$$

  Using the countable axiom of choice we can construct a subsequence ${(f_{N_i})}_{i\in \mathbb{N}}$ such that $\forall <!--\; FUNCTION\; APPLICATION\; -->N_i\in \mathbb{N}$ and $\forall <!--\; FUNCTION\; APPLICATION\; -->x\in {ℝ}^d$ we have $d(f_{N_i}(x),f(x))\le 1∕i$. Since $f_n$ are increasing, we also have this property for all $n>N_i$. In other words,

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->\frac{1}{i}>0\exists <!--\; FUNCTION\; APPLICATION\; -->N_i\in \mathbb{N}\forall <!--\; FUNCTION\; APPLICATION\; -->n>N_i:d_{\infty }(f_n,f)\le \frac{1}{i}$$

  as desired.

For the backward implication, I follow the solution by Yaver Gulusoy.

For the forward implication, define $M=\underset{x\in {ℝ}^d}{\sup <!--\; FUNCTION\; APPLICATION\; -->}f(x)$. Consider a sequence of simple functions ${\{f_n\}}_{n\in \mathbb{N}}$ defined by

Where $\lfloor 𝜃\rfloor$ is the "floor" of $𝜃$, the greatest integer less than or equal to $𝜃$.

This functions in this sequence are simple because for each $n\in \mathbb{N}$, $f_n$ only can take on the values between $0$ and $M$ in steps of $M∕2^n$.

I claim that this sequence converges uniformly to $f$. Indeed, for $𝜖>0$, let $N$ be such that $M∕2^N\le 𝜖$. Then for $n>N$ and $x\in {ℝ}^d$ we have

Proof

Let $f:{ℝ}^d\to [0,+\infty ]$. We aim to show that $f$ is a bounded unsigned measurable function if and only if $f$ is the uniform limit of bounded simple functions.

Construction of an Exemplary Sequence $f_n$

By $f$ being bounded, we have that $A\le f(x)\le B$. Let $n\in \mathbb{N}$ and partition $[A,B]$ into $n$ half-open slices of width

with $A_i=[A+(i-1)M,A+\mathit{iM})$ for $1\le i<n$ and $A_n=[A+(n-1)M,B]$.

It holds that

Using the property of $f$, it follows that

because for each $x\in {ℝ}^d$, the property $A\le f(x)\le B$ implies that $f^{-1}([A,B])$ covers all of the domain. Hence, ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{f}^{-1}(A_i)$ is a partition of ${ℝ}^d$. Name $f^{-1}(A_i)=E_i$.

Now, denote the lower point of each $A_i$ as

and set

By MTP3.9(9), $E_i$, as the preimage of an interval, is Lebesgue measurable.

Convergence Analysis

For any $x\in {ℝ}^d$, $x$ must also belong to some $f^{-1}(A_i)$, and thus:

Since $x\in f^{-1}(A_i)$, this simplifies to:

where $f(x)\ge c_n^i$. Hence:

because $f(x)\le A+\mathit{iM}$ and $c_n^i=A+(i-1)M$. Subtracting these, we have:

Now consider an arbitrary $𝜀>0$ and let $n\in \mathbb{N}$ be the ceiling integer for $\frac{|B-A|}{𝜀}$. Then:

For an arbitrary $x\in {ℝ}^d$, let $x\in f^{-1}(A_i)$ for some $1\le i\le n$. Then:

Hence, $f_n$ converges uniformly to $f$. □