It is seen that, if $f(x)$ is a bounded real valued function, then $\displaystyle \overline{\int_a^b } f(x)\; dx= \inf \int_a^b \psi(x) dx \; $ where infimum is taken for all step functions $\psi(x)\ge f(x)$. Why we take all step functions $\; \psi(x)\ge f(x)$
Riemann Integral in terms of step function
0
$\begingroup$
integration
riemann-integration
-
0Why not? What would you propose instead? – 2017-02-27
-
0@JoshuaRuiter Sir I would like to know why we take step functions – 2017-02-27
-
0@JoshuaRuiter we already defined upper integral as $\inf S$ where S=\Sum_{i=1}^n M_i (x_i-x_{i-1}$ where $M_i= Sup f(x) ,x\in [x_{i-1},x_i]$ – 2017-02-27
-
1@SeenaV it's because we already know how to integrate step functions, and it is quite easy, whereas we have not defined how to integrate arbitrary functions. – 2017-02-27