Question about Riemann integral property proof

Question

I just read the following property and proof:

Let $I = [a_1, b_1] \times ... \times [a_n, b_n]\subset \mathbb{R}^n$ and $f,g : I \rightarrow \mathbb{R}$ integrable functions such that $f \leq g$. Then $\int_I f \leq \int_I g$.

Proof: It is a consequence of the property $U(f,P) \leq U(g,P)$.

I don' t really understand it.

Because $f,g$ are integrable, then its upper Riemann integrals coincide with its integrals, so $\int_I f \leq \int_I g \iff \inf \{U(f,P): \text {$P$ partition of $I$} \} \leq \inf \{U(g,P): \text {$P$ partition of $I$} \}$.

But in the proof it is assumed that the partition $P$ that makes $U(f, P)$ the smallest is the same that the one that makes $U(g,P)$ the smallest. Why is this true? Or does it mean that $U(f,P') \leq U(g,P'')$ for any $P', P''$ partitions of $I$ ?

Answer 1

First of all, there is no reason for a partition that makes $U(f, P)$ the smallest to exist.

Second of all, the assumption you mention is unnecessary. For each $\varepsilon > 0$ there is a partition $P$ such that $U(g, P) \leqslant \int_I g + \varepsilon$ and then

$$\int_I f \leqslant U(f, P) \leqslant U(g, P) \leqslant \int_I g + \varepsilon$$

Since $\varepsilon > 0$ is arbitrary, we get $\int_I f \leqslant \int_I g$.

This fact is an example of a more general one: let $A$ and $B$ be non-empty, bounded from below sets of real numbers. If for each $b \in B$ there is an $a \in A$ such that $a \leqslant b$, then $\inf A \leqslant \inf B$. Now taking $A = \{ U(f, P) : P \text{ is a partition of } I \}$ and $B = \{ U(g, P) : P \text{ is a partition of } I \}$, we get the desired result.

Answer 2

It is the same to prove that if $$f\le 0$$ then $$J=\int f\le 0$$.

assume that $J>0$.

then with $\epsilon=\frac{J}{2}$, there exist a step $p>0$ such that for all subdivision $\sigma$ with $|\sigma|<p$, the Riemann sums satisfy

$$-\frac{J}{2}<\sum (x_{i+1}-x_i)f(\xi_i)-J<\frac{J}{2}$$

but all Riemann sums are nonpositive.