Prove if f is positive and increasing on $[a, b]$ then $L_n ≤ A ≤ R_n.$ (riemann sum)

Prove if f is positive and increasing on $[a, b]$ then for all $n\ge 0$ we have $L_n \le A \le R_n$. (Riemann sum)

Let $A$ denote the actual area.

Let $L_n$ denote the left Riemann sum.

Let $R_n$ denote the right Riemann sum.

So far what I did:

$$\int_{a}^{b} f(a) dx \leq \int_{a}^{b} f(x) dx \leq \int_{a}^{b} f(b)$$

But after this I am stuck on the proof?

Just write the definitions

$$L_n = \sum_{i=0}^{n-1}f(x_i){b-a\over n}\le \sup_m L_m = A= \inf_m R_m \le \sum_{i=1}^n f(x_i){b-a\over n}$$

here the $\sup=\limsup$ by monotonicity of $f$.

and if you're not familiar with the $\sup$ notation, then try this

$$L_n =\sum_{i=0}^{n-1}f(x_i){b-a\over n}=\sum_{i=0}^{n-1}\int_{x_i}^{x_{i+1}}f(x_i)\,dx \le \sum_{i=1}^{n-1}\int_{x_i}^{x_{i+1}}f(x)\,dx$$ $$= \int_a^b f(x)\,dx$$ $$=\sum_{i=1}^n\int_{x_{i-1}}^{x_i}f(x)\le \sum_{i=1}^n\int_{x_{i-1}}^{x_i}f(x_i)\,dx =\sum_{i=1}^nf(x_i){b-a\over n}=R_n$$

• Does the downvoter want to explain what he/she thinks is wrong with my answer? – Adam Hughes Jan 10 '17 at 0:22
• Is the first claim clearly true? I can see that $L_n \le \sup_m L_m$, and for an integrable $f$, $\limsup_m L_m = A$, but I think that going from $\sup_m L_m$ to $\limsup_m L_m$ possibly requires a little more justification. (same comment w.r.t. the upper bound). – πr8 Jan 10 '17 at 0:47
• @πr8 monotonicity means $\limsup = \sup$. I've edited this detail in. – Adam Hughes Jan 10 '17 at 0:51
• The function is monotone, but is it clear that the upper/lower sums are monotone as a result of this? Will change to an upvote in any case - was hasty to downvote, and the remainder of your answer is exemplary. – πr8 Jan 10 '17 at 1:01

For partition $x_0<x_1<\cdots<x_n$

Left Riemann sum: $= \sum_\limits{i=1}^{n} f(x_{i-1})(x_{i}-x_{i-1})$

Right Riemann sum: $= \sum_\limits{i=1}^{n} f(x_i)(x_i-x_{i-1})$