Riemann Integral: prove $\{U(P,f):P$ is a partition of $[a,b]\}$ is bounded below

Question

[Defining the Riemann Integral: ]

We consider a partition, $P=\{a=x_0<x_1<...<x_n=b\}$ of $[a,b]$ and a bounded function $f:[a,b]\rightarrow\mathbb{R}$.
Next, we define- $$M_i=\sup\{f(x)|x\in[x_{i-1},x_i]\}$$ and $$U(P,f)=\underset{i=1}{\overset{n}{\sum}}M_i(x_i-x_{i-1})$$

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What was also mentioned was $A_1=\{U(P,f):P$ is a partition of $[a,b]\}$ is bounded below.
Intuitively, of course, the actual area covered by $f(x)\le A_1$, which is why we can say that $A_1$ is bounded below. But, how do I formally prove that?

Answer 1

Suppose $P$ is a partition and that $Q$ is the partition obtained by $P$ by adding a single point. Prove that $L(P,f)\le L(Q,f)$, where $L$ denotes the lower sum.

Since any partition $P$ can be obtained from the trivial partition $P_0=\{a,b\}$ by successively adding points, induction shows that $L(P_0,f)\le L(P,f)$.

Next observe that $L(P,f)\le U(P,f)$.

Similarly, lower sums are upper bounded.

It is perhaps simpler to observe that $$ \sum_{i=1}^n M_i(x_i-x_{i-1})\ge\sum_{i=1}^n m(x_i-x_{i-1})=m(b-a) $$ where $m=\inf\{f(x):x\in[a,b]\}$, but the above approach allows for going on with the construction of the Riemann integral.

Answer 2

egreg's answer is excellent (+1), but I thought I'd elaborate on my comment.

We are given that $f$ is bounded, which by definition means that there exists $M\in\Bbb{R}$, $M>0$ such that $$|f(x)|<M$$ for $x\in[a,b]$.

Then $f(x)>-M$ for all $x\in[a,b]$, and therefore $U(P,f) > -M(b-a)$.