Of course you need calculus to buy groceries (Measure, Integration and Grocery/Amazon Shopping)

We know that there is a saying in life that mathematics is useless. Do you need calculus to buy groceries? As a mathematics student and someone who goes to the supermarket every other day to buy groceries and cook, my answer is that it seems essential, not only in the supermarket but also in Amazon.

But we can not use Riemann integral to do this; that is why the Riemann integral is bad and why we need Measure Theory (just kidding.

What is Measure space?

Measure space is a triple $(X,\mathrm{\Sigma },\mu )$$(X,\Sigma,\mu)$, $X$$X$ is a set, $\mathrm{\Sigma }$$\Sigma$ is a sigma-algebra, and $\mu$$\mu$ is a measure function.

A σ-Algebra is a kind of Boolean algebra, it requires closure under complementation and countable union operations. By De Morgan’s law, this implies closure under a countable intersection.

Because $\begin{array}{c}\bigcap _{i=1}^{\mathrm{\infty }}{A}_i={(\bigcup _{i=1}^n(A_i^c))}^c\end{array}$$\bigcap_{i=1}^\infty A_i=\left(\bigcup_{i=1}^n(A_i^c)\right)^c\\$

This, in turn, implies that $\mathrm{\varnothing }=A\cap A^c\in \mathrm{\Sigma },X=A\cup A^c\in \mathrm{\Sigma }$$\O=A\cap A^c\in\Sigma,X= A\cup A^c\in\Sigma$

Measure function $\mu :\mathrm{\Sigma }\to {\mathbb{R}}^{+}$$\mu:\Sigma\to\mathbb R^+$ have properties as follows

$\mu (\mathrm{\varnothing })=0$$\mu(\O)=0$

$\begin{array}{c}\mu (\underset{i=1}{\overset{\mathrm{\infty }}{⨆}}{A}_i)=\sum _{i=1}^{\mathrm{\infty }}\mu (A_i)\end{array}$$\mu(\bigsqcup_{i=1}^\infty A_i)=\sum_{i=1}^\infty\mu(A_i)\\$, $\bigsqcup$$\sqcup$ is disjoint union

View supermarkets and Amazon as Measure space

Consider all the goods in supermarkets or Amazon

Let $\mathrm{\Sigma }:=P(X)$$\Sigma:=P(X)$, and $\mu$$\mu$ be the price.

Check is a Measure space! ( also supermarket)

When we check out, what we do actually is, integral for nonnegative simple functions!

The nonnegative simple functions is such as $\begin{array}{c}f(x)=\sum _{i=1}^n{\lambda }_i{\chi }_{A_i}(x)\end{array}$$f(x)=\sum_{i=1}^n \lambda_i\chi_{A_i}(x)\\$

$\begin{array}{c}{\chi }_{A_i}(x)=\{\begin{array}{l}1,x\in A_i\\ 0,x\notin A_i\end{array}\end{array}$$\chi_{A_i}(x)=\begin{cases}1,x\in A_i\\0,x\notin A_i\end{cases}$

In this case, it is just whether you have bought it or not.

OK, so, what is integral for nonnegative simple functions on a Measure space?

The definition is $\begin{array}{c}{\int }_{X}fd\mu =\sum _{i=1}^n{\lambda }_i\mu (A_i)\end{array}$$\int_X fd\mu=\sum_{i=1}^n\lambda_i\mu(A_i)\\$

And $\begin{array}{c}f(x)=\sum _{i=1}^n{\lambda }_i{\chi }_{A_i}(x)\end{array}$$f(x)=\sum_{i=1}^n \lambda_i\chi_{A_i}(x)\\$ represents what you bought, it is just like a list.

For example, what you want is $\begin{array}{c}A=\bigcup _{i=1}^n\{a_i\}\end{array}$$A=\bigcup_{i=1}^n\{a_i\}\\$

And $\begin{array}{c}f(x)=\sum _{i=1}^n{\lambda }_i{\chi }_{\{a_i\}}(x)\end{array}$$f(x)=\sum_{i=1}^n \lambda_i\chi_{\{a_i\}}(x)\\$, ${\lambda }_i$$\lambda_i$ is how many pieces of $a_i$$a_i$ you buy

When you check out, the money you need pay is just $\begin{array}{c}{\int }_{X}fd\mu =\sum _{i=1}^n{\lambda }_i\mu (\{a_i\})\end{array}$$\int_X fd\mu=\sum_{i=1}^n\lambda_i\mu(\{a_i\})\\$

For example, in $(\mathrm{Amazon},\mathrm{\Sigma },u)$$(\mathrm{Amazon},\Sigma,\,u)$, the integral for the nonnegative simple function is just like