The proof of Theorem 3 on page 73 in Royden "fourth edition".

Question

The theorem and its proof are given below:

My questions are:

1- I do not understand the last part of the last statement in the proof "Also the equality of the Riemann and Lebesgue integrals", how what is mentioned before this in the proof leads to the equality of the Riemann and Lebesgue integrals?

2- When my professor proved this theorem She/He used the following inequalities:

$$(R) \int_{-} \leq (L)\int_{-}f \leq (L)\int^{-}f \leq (R)\int^{-}f ,$$ where $R$ stands for Riemann integrable and $L$ stands for Lebesgue integrable and the dashes above and below the integration sign stands for upper and lower integrals respectively.

My question is I do not understand why the upper Riemann integral should be the largest one? why Lebesgue integrals should be in between Riemann integrals? why the lower Riemann integral should have the least value?

Could anyone help me answer those questions please?

Answer 1

Because

• each step function is a simple function,
• for each step function the Riemann integral and the Lebesgue integral are the same,
• for any subsets $A$ and $B$ of real numbers, if $A\subset B$, then $\sup A\le\sup B$ and $\inf A\ge \inf B$.

one has $\sup A_R\le \sup A_L$ and $\inf B_R\ge \inf B_L$, where $$ A_R:= \left\{ (R)\int_I\phi \mid \phi \text{ is a step function, }\phi\le f \right\}, $$ $$ A_L:= \left\{ \int_I\phi \mid \phi \text{ is a simple function, }\phi\le f \right\}, $$ $$ B_R:= \left\{ (R)\int_I\phi \mid \phi \text{ is a step function, }f\le \phi \right\}, $$ $$ B_L:= \left\{ \int_I\phi \mid \phi \text{ is a simple function, }f\le \phi \right\}. $$

So it follows that $$ \inf B_L\le \inf B_R=\sup A_R\le \sup A_L. $$ But $\inf B_L=\sup A_L$. Therefore these four quantities must be the same: $$ \inf B_L= \inf B_R=\sup A_R= \sup A_L. $$