I have to show that the $L^1$ norm of the Dirichlet function is equal to zero. I know that the $L^1$ norm has the formula:
$$||D(x)||_1 = \big(\int |D(x)| dx\big)$$
and the Dirichlet function is:
$$D(x) = \begin{cases} 1 & x \in \mathbb{Q} \\ 0 & x \in\mathbb{I} \end{cases}$$
On
You need to interpret the $L_1$ norm of an element as a Lebesgue integral, not as a Riemann integral since the latter one does not even exist for this particular function. Indeed $$\overline{\int_{\Omega}}D(x)\,dx=\inf_{P\in\mathcal{P}}U(D,P)=\inf_{P\in\mathcal{P}}\sum_{i=1}^n M_iv(I_i)=\inf_{P\in\mathcal{P}}\sum_{i=1}^nv(I_i)=v(\Omega)$$ and $$\underline{\int_{\Omega}}D(x)\,dx=\sup_{P\in\mathcal{P}}U(D,P)=\sup_{P\in\mathcal{P}}\sum_{i=1}^n m_iv(I_i)=\inf_{P\in\mathcal{P}}\sum_{i=1}^n0\cdot v(I_i)=0\neq v(\Omega)$$ where $M_i:=\max_{x\in I_i}D(x)=1$ and $m_i:=\min_{x\in I_i}D(x)=0$ and $\mathcal{P}$ is the set of all partitions of the domain $\Omega$ (note that $\Omega=\mathbb{R}$ possibly). Hence the lower and upper itegrals don't coincide. This implies $D(x)$ is not Riemann integrable. On the other hand as a Lebesgue integral it makes sense since $$\int_{\Omega}|D(x)|\,dx:=\sup_{f\in S}\{\int f\,dx: f(x)\leqslant |D(x)|,\forall x\in\Omega\}$$ where $S$ is the set of all simple functions (expressible as a linear combination of step functions). From the definition of $D(x)$ the only simple functions $f$ that are dominated by $D$ are the ones that are non-positive everywhere. Otherwise there would exists some interval $I\subseteq \Omega$ such that $f(x)=c>0$. But irrationals are dense in $\mathbb{R}$ and so there would exists some irrational number $y\in I$ implying $D(y)=0<c=f(y)$. This would contradict $f(x)\leqslant D(x)$ for all $x$. It follows now that $$\sup_{f\in S}\{\int f\,dx: f(x)\leqslant |D(x)|,\forall x\in\Omega\}=0\Rightarrow ||D||_{L_1}=\int_{\Omega}|D(x)|\,dx=0$$
This $L^1$ norm equal to the (Lebesgue) measure of $\mathbb{Q}$, which is zero.