What if we take step functions instead of simple functions in the Lebesgue integral

Question

When we define the Lebesgue integral, we first define it for simple functions $s(x) = \sum\limits_{j=1}^n c_j\chi_{A_j}(x)$ (where $A_j$ are measurable) as $\int sd\mu = \sum\limits_{i=j}^n c_j \mu(A_j)$ and then for $f\ge 0$ as $\int fd\mu = \sup\{\int sd\mu$ : s simple and $0\le s\le f\}$. But I was wondering what could go wrong if instead of taking simple functions in the supremum, we would take step functions, i.e. $s(x)=\sum\limits_{j=1}^nc_i\chi_{I_j}(x)$ where $I_j$ are intervals (any type, like $(a,b), (a,b], [a,b), [a,b])$).

Answer 1

It is easy to approximate measurable functions by simple functions: There is a 'simple'construction of a sequence of simple functions increasing to a non-negative measuarble function. But it is not even true that any measuarble funciton is the ponit-wise limit of a sequence of step functions. However, it can be shown that if $f$ is integrable and $\epsilon >0$ then there exists a step function $g$ with $\int |f-g|d\mu <\epsilon$.

Answer 2

Your question: What could go wrong if instead of taking simple functions in the supremum, we would take step functions, i.e. $s(x)=\sum\limits_{j=1}^nc_i\chi_{I_j}(x)$ where $I_j$ are intervals (any type, like $(a,b), (a,b], [a,b), [a,b])$)?

Answer: If the Integral was defined using step functions, it would not be additive. Note:

$$ \int \chi_{ \mathbb{Q}\cap[0,1]} = \sup\{\int sd\mu : \textrm{ s step function and } 0\le s\le \chi_{\mathbb{Q})\cap[0,1]}\}=0 $$

$$\int \chi_{(\mathbb{R} \setminus \mathbb{Q})\cap[0,1]} = \sup\{\int sd\mu : \textrm{ s step function and } 0\le s\le \chi_{(\mathbb{R} \setminus \mathbb{Q})\cap[0,1]}\}=0 $$

$$ \int \chi_{ [0,1]} = \sup\{\int sd\mu : \textrm{ s step function and } 0\le s\le \chi_{[0,1]}\}=1 $$

So $\chi_{ [0,1]} = \chi_{ \mathbb{Q})\cap[0,1]} + \chi_{(\mathbb{R} \setminus \mathbb{Q})\cap[0,1]}$ , but $$ \int \chi_{ [0,1]} = 1 \neq 0 + 0 = \int \chi_{ \mathbb{Q}\cap[0,1]} + \int \chi_{(\mathbb{R} \setminus \mathbb{Q})\cap[0,1]} $$

The only way to recover additivity would be to reduce the class of integrable functions. So either Integrals would not be additive or we would have less integrable functions.