Measurable Functions and Simple Approximation - Examples

The set $E$ is measurable by hypothesis.

Therefore [L1] gives that ${\mathbf{1}}_E$ is measurable.

The function $s$ takes only the three values $2$, $-1$, and $0$, and the [given] corresponding level sets are exactly the three Borel sets displayed above. Thus $s$ is measurable.

Therefore [L1] identifies $s$ as a simple function and the displayed sum as [step 1.1, L1] its canonical representation.

By [L1], the function $D$ is Borel measurable.

Let $x\in \mathbb{R}$ and let $U$ be any neighbourhood of $x$. By [L2], the [step 1.1, given, L2] set $U$ contains both a rational point and an irrational point, so $D$ takes both values $1$ and $0$ on $U$. Therefore $D$ cannot be continuous at $x$. Since $x$ was arbitrary, $D$ is nowhere continuous.

By [L1], the Cantor function is continuous on $[0,1]$.

Applying [L2] to the continuous map $c:[0,1]\to \mathbb{R}$ gives that [step 1.1, L2] $c$ is Borel measurable on the subspace $[0,1]$.

For $k=1$, the two dyadic cells are [L1, algebra] $0\le x^2<1\mathrm{/}2$ and $1\mathrm{/}2\le x^2<1$, while the truncation cell is $x^2\ge 1$. On $[0,2]$ these become exactly the three intervals displayed in the first formula.

For $k=2$, the eight nontrivial dyadic cells are [step 1.1, L1, algebra] $j\mathrm{/}4\le x^2<(j+1)\mathrm{/}4$ for $j=0,\dots ,7$, together with the truncation cell $x^2\ge 2$. On $[0,2]$ these become $[0,1\mathrm{/}2)$, $[1\mathrm{/}2,\sqrt{2}\mathrm{/}2)$, $[\sqrt{2}\mathrm{/}2,\sqrt{3}\mathrm{/}2)$, $[\sqrt{3}\mathrm{/}2,1)$, $[1,\sqrt{5}\mathrm{/}2)$, $[\sqrt{5}\mathrm{/}2,\sqrt{6}\mathrm{/}2)$, $[\sqrt{6}\mathrm{/}2,\sqrt{7}\mathrm{/}2)$, $[\sqrt{7}\mathrm{/}2,\sqrt{2})$, and $[\sqrt{2},2]$, exactly as displayed.

Because $f$ takes only the values $0$ and $1$, every preimage $f^{-1}(B)$ is [L1, given] one of $\mathrm{\varnothing }$, $X$, $A$, or $X\setminus A$. So $\sigma (f)$ is contained in that four-set family.

The four sets all occur as preimages: [step 1.1, L1] $A=f^{-1}((1\mathrm{/}2,\mathrm{\infty }))$, $X\setminus A=f^{-1}((-\mathrm{\infty },1\mathrm{/}2))$, $X=f^{-1}(\mathbb{R})$, and $\mathrm{\varnothing }=f^{-1}(\mathrm{\varnothing })$. Therefore they are exactly the members of $\sigma (f)$.

Because $f$ takes only the two values $\alpha$ and $\beta$, every preimage [L1, given] $f^{-1}(B)$ is one of $\mathrm{\varnothing }$, $X$, $A$, or $X\setminus A$. So $\sigma (f)$ is contained in that four-set family.

Since $\alpha \mathrm{\ne }\beta$, choose a real threshold strictly between them. The [step 1.1, L1] corresponding open ray has preimage $A$ or $X\setminus A$, and the other three sets arise as in the previous example from $\mathbb{R}$ and $\mathrm{\varnothing }$. Hence all four sets belong to $\sigma (f)$, so they are exactly its members.

Let $f:={\mathbf{1}}_E$. Because $E$ is Lebesgue measurable, [L2] makes $f$ [L1, L2] Lebesgue measurable.

But

Every singleton $\{t\}$ belongs to the trace Lebesgue sigma-algebra, so [L2] makes each ${\mathbf{1}}_{\{t\}}$ measurable.

Their pointwise supremum is ${\mathbf{1}}_V$. If $V$ belonged to the trace Lebesgue sigma-algebra, then $V=E\cap [0,1]$ for some Lebesgue measurable $E\subseteq \mathbb{R}$, making $V$ Lebesgue measurable in $\mathbb{R}$, contrary to [L1]. Thus [L2] makes ${\mathbf{1}}_V$ nonmeasurable, so the arbitrary-family version fails.

By [L1], the set $E$ is Lebesgue measurable and the map $g$ is continuous. [L1]

The same fact [L1] says that $g^{-1}[E]$ is not Lebesgue measurable, which [step 1.1, L1] is exactly the required failure.

Let $f=0$ and $g={\mathbf{1}}_E$. The set $C$ is Borel null, so $g=f$ on [L1] $\mathbb{R}\setminus C$ and hence $g=f$ almost everywhere.

But [step 1.1, L1] $g^{-1}((1\mathrm{/}2,\mathrm{\infty }))=E$, and [L1] says $E$ is not Borel. Thus $g$ is not measurable for the Borel sigma-algebra, even though it agrees almost everywhere with the measurable function $f$.

Every level set of $f$ is empty, a singleton, or a two-point set, so every [given] level set is measurable.

Yet $f^{-1}([2,3])=V$, and [L1] says that $V$ is not measurable. Therefore [step 1.1, L1] $f$ is not measurable.