Why is $\mathbb{1}_{\{0\}} = \mathbb{1}_\mathbb{Q} = 0$ and $\mathbb{1}_{[0,1]\backslash\mathbb{Q}} = \mathbb{1}_{[0,1]}$ in $L^p(\mathbb{R})$? I have the following definition, where $\mathbb{K} = \mathbb{R}$ or $\mathbb{K} = \mathbb{C}$:
Definition: For $p\in [1,\infty)$ let $$L^p(S) = \{f:S\to\mathbb{K}:\text{ $f$ is measurable and }\int_S \left|f\right|^p d\mu <\infty\}.$$
For $f\in L^p(S)$ let $$\left\Vert f\right\Vert_p:= \bigg(\int_S\left|f\right|d\mu\bigg)^{\frac{1}{p}}$$
The book I'm using then gives an example and states that if $S = \mathbb{R}$ with the Lebesgue measure $\lambda$ we have that $\mathbb{1}_{\{0\}} = \mathbb{1}_\mathbb{Q} = 0$ and $\mathbb{1}_{[0,1]\backslash\mathbb{Q}} = \mathbb{1}_{[0,1]}$ in $L^p(\mathbb{R})$.
Question: Why is this example true? What does it even mean that $\mathbb{1}_{[0,1]\backslash\mathbb{Q}} = \mathbb{1}_{[0,1]}$ in $L^p(\mathbb{R})$?
Thanks in advance!
 A: Stated another way.  To make $L^p$ a genuine Banach space, we need $$\|f\|_p = 0 \quad\Longrightarrow\quad f=0$$
This fails for actual functions, so we can use a quotient space by all functions that vanish almost everywhere.  Thus, two functions are in the same coset if they agree almost everywhere.  But that is cumbersome, so we just remember that convention and talk about actual functions.  So that is the sense of equality used in the question.
It is not true that
$$
\mathbb{1}_{[0,1]\backslash\mathbb{Q}} = \mathbb{1}_{[0,1]}
$$
for actual functions.  But these two functions belong to the same coset, so they are (or they define) the same element of $L^p$.  
I remember the textbook of Hewitt & Stromberg use two different different notations, maybe $L^p$ and ${\scr{L}}^p$, for the space of functions and the space of equivalance classes.  Then (after a certain point) they revert to just one notation.
A: What they are essentially saying is that two functions are equal if they only differ for sets of measure 0 in their domain. For Lebesgue measure, sets of countably many points have measure 0. So, for instance, the indicator for $\mathbb{Q}$ differs from the 0 function only on rationals, which are countable. As such, $\mathbb{1}_\mathbb{Q} = 0$.
