Show is that the Riemann integral $\int_a^b f(x)\,dx$ is bounded by $M(b-a)$, where $M$ is a bound of $f(x)$.
I was thinking I would show that $M(b-a)$ is larger than every Riemann sum, but my proof includes the phrase "choose a tagged partition such that the absolute value of the associated Riemann sum is larger than the absolute of the integral. But I don't know if this is always possible.
Can you give me a hint as to how to show this, if it is possible? I need it for a homework problem I'm working on, and I'm technically not supposed to know about the Upper Riemann sum of a function, although I've already gone through the trouble of defining it.
Hint: Take the Riemann sum where the partitioning of $[a,b]$ is simply $\{[a,b]\}$ (in other words, $x_0=a, x_1=b$), and pick the element from $[a,b]$ for which the Riemann sum will be large enough.
EDIT:
To prove that the integral $\int_a^bf(x)dx$ is bounded by $M(b-a)$, where $M$ is a bound for $f$, it is enough to show that for every Riemann sum $S$, you have $S<M(b-a)$ (since if it is true for every Riemann sum, it must be true for their limit as well).
This is fairly simple, since for some $a=x_0<x_1<x_2<\dots<x_n=b$ and $x_k^*\in[x_{k-1},x_k]$ for $k=1,2\dots, n$, you know that $$S=(x_1-x_0)f(x_1^*) + (x_2-x_1)f(x_2^*) + \cdots + (x_n-x_{n-1})f(x_n^*)$$
Now, use the fact that for every $k$, you have $f(x_k^*)<M$. What equality do you get?