The Dirichlet function is the pointwise limit of a sequence of Baire class one functions and is itself not Baire class one, so the Baire hierarchy on $[0,1]$ is already strict at the first level

Fix a surjection $s:\mathbb{N}\to \mathbb{Q}\cap [0,1]$ and, for $m\in \mathbb{N}$, let $g_m:[0,1]\to \mathbb{R}$ be the indicator of $\{s(0),\dots ,s(m)\}$: $g_m(x)=1$ if $x=s(j)$ for some $j\le m$, and $g_m(x)=0$ otherwise.

Suppose, for contradiction, that $D$ is of Baire class one on $[0,1]$.

But $D$ is continuous at no point of $[0,1]$. Let $c\in [0,1]$ and let $\delta >0$ be real; put $u:=\max \{0,c-\delta \mathrm{/}2\}$ and $v:=\min \{1,c+\delta \mathrm{/}2\}$, so that $u<v$ and $[u,v]\subseteq [0,1]\cap N_{\delta }(c)$, the strict inequality holding because $c\in [0,1]$, $0<1$ and $\delta >0$. The nondegenerate interval $(u,v)$ contains a rational $y_1$ and an irrational $y_2$, with $D(y_1)-D(y_2)=1$, so one of $\mathrm{\mid }D(y_1)-D(c)\mathrm{\mid }$ and $\mathrm{\mid }D(y_2)-D(c)\mathrm{\mid }$ equals $1$; hence no $\delta$ witnesses continuity at $c$ for $\epsilon =1$. This is the argument of the Dirichlet claim, restricted to the domain $[0,1]$.

For $m\in \mathbb{N}$ define ${\rho }_m:[0,1]\to \mathbb{R}$ by ${\rho }_m(x):=\min \{\mathrm{\mid }x-s(j)\mathrm{\mid }:j\le m\}$, the minimum of a nonempty finite set of reals. Then ${\rho }_m(x)\ge 0$, and ${\rho }_m(x)=0$ exactly when $x=s(j)$ for some $j\le m$, the minimum being attained.

Then the set of points of $[0,1]$ at which $D$ is continuous is dense in $[0,1]$, since $0<1$; in particular it is nonempty.

${\rho }_m$ is $1$-Lipschitz, hence continuous: choosing $j_0\le m$ with ${\rho }_m(x)=\mathrm{\mid }x-s(j_0)\mathrm{\mid }$ gives ${\rho }_m(y)\le \mathrm{\mid }y-s(j_0)\mathrm{\mid }\le \mathrm{\mid }y-x\mathrm{\mid }+{\rho }_m(x)$, and exchanging $x$ and $y$ gives $\mathrm{\mid }{\rho }_m(x)-{\rho }_m(y)\mathrm{\mid }\le \mathrm{\mid }x-y\mathrm{\mid }$; so $\delta :=\epsilon$ witnesses continuity at every point.

For $m,n\in \mathbb{N}$ define $h_{m,n}:[0,1]\to \mathbb{R}$ by $h_{m,n}(x):=\max \{0,1-\iota (n){\rho }_m(x)\}$; the index runs over the whole of $\mathbb{N}$, the term at $n=0$ being the constant $1$ since $\iota (0)=0$, so that $n\mapsto h_{m,n}$ is a sequence in the sense of Sequences of reals: bounded, eventually, frequently, tails, subsequences. Each $h_{m,n}$ is continuous on $[0,1]$, being the pointwise maximum of the constant $0$ and the continuous function $x\mapsto 1-\iota (n){\rho }_m(x)$.

For each fixed $m$ the sequence $n\mapsto h_{m,n}$ converges pointwise on $[0,1]$ to $g_m$. If ${\rho }_m(x)=0$ then $h_{m,n}(x)=\max \{0,1\}=1=g_m(x)$ for every $n$. If ${\rho }_m(x)>0$ then, taking a natural $n_0\ge 1$ with $\iota (n_0)>1\mathrm{/}{\rho }_m(x)$, every $n\ge n_0$ has $\iota (n){\rho }_m(x)\ge \iota (n_0){\rho }_m(x)>1$, so $h_{m,n}(x)=0=g_m(x)$.

Hence each $g_m$ is of Baire class one on $[0,1]$, being the pointwise limit of a sequence of continuous functions.

The sequence $(g_m{)}_{m\in \mathbb{N}}$ converges pointwise on $[0,1]$ to $D$. If $x\in \mathbb{Q}\cap [0,1]$ then $x=s(k)$ for some $k$, since $s$ is onto, and $g_m(x)=1=D(x)$ for every $m\ge k$. If $x\in [0,1]$ is irrational then $x\mathrm{\ne }s(j)$ for every $j$, so $g_m(x)=0=D(x)$ for every $m$. Claim 1 is proved.

Steps 2.2 and 1.3 contradict one another, so the assumption of step 1.2 is false and $D$ is not of Baire class one on $[0,1]$: claim 2 holds, and with step 7.1 the example is complete.