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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?

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    didn't you ask this question here: http://math.stackexchange.com/questions/2089043/how-to-prove-left-riemann-sum-is-underestimate-and-right-riemann-sum-is-overesti/2089049?noredirect=1#comment4294703_2089049 ? – 2017-01-10
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    I did but i just wanted more answers, so i asked again – 2017-01-10

2 Answers 2

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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

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

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    Does the downvoter want to explain what he/she thinks is wrong with my answer? – 2017-01-10
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    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). – 2017-01-10
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    @πr8 monotonicity means $\limsup = \sup$. I've edited this detail in. – 2017-01-10
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    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. – 2017-01-10
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For partition $x_0

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})$