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Q) For a given function on a given interval, the difference between the left-hand sum and right-hand sum always gets smaller as the number of subdivisions gets larger.

I remember that the answer is True but I forgot the reason.

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Are you looking for a rigorous answer, or a "hand-waving" answer? –  anorton Jul 9 at 2:10
    
@anorton I need a concise explanation :) –  John Jul 9 at 2:16
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Whether true or false depends on how your book define the concept. It sounds like it's actually false here. Looks at Wikipedia for a more detailed answer: en.wikipedia.org/wiki/Riemann_integral#Similar_concepts –  Gina Jul 9 at 6:50

4 Answers 4

Are we allowed to have badly discontinuous functions and irregular subdivisions?

If so, let's look at left and right endpoint sums for $\displaystyle\int_{0}^{1}f(x)\,dx$ where $f(x) = \begin{cases}1 & x \in \mathbb{Q} \\ 0 & x \not\in \mathbb{Q}\end{cases}$.

Subdivide $[0,1]$ into $2n$ intervals: $[x_0,x_1], \ldots, [x_{2n-1},x_{2n}]$ where $x_{2k} = \dfrac{k}{n}$ and $x_{2k+1} = \dfrac{k+1/\pi}{n}$.

Clearly, $x_k - x_{k-1} = \dfrac{1/\pi}{n}$ if $k$ is odd and $\dfrac{1-1/\pi}{n}$ if $k$ is even.

Also, $x_{2k} \in \mathbb{Q}$ and $x_{2k+1} \not\in \mathbb{Q}$. So, $f(x_k) - f(x_{k-1}) = -1$ if $k$ is odd and $1$ if $k$ is even.

Then, the difference between the left and right endpoint sums is:

$R-L = \displaystyle\sum_{k = 1}^{2n}f(x_k)(x_{k}-x_{k-1}) - \sum_{k = 1}^{2n}f(x_{k-1})(x_{k}-x_{k-1}) = \sum_{k = 1}^{2n}(f(x_k)-f(x_{k-1}))(x_{k}-x_{k-1}) = \sum_{\substack{k = 1 \\ k \ \text{is odd}}}^{2n}-1 \cdot \dfrac{1/\pi}{n} + \sum_{\substack{k = 1 \\ k \ \text{is even}}}^{2n}1 \cdot \dfrac{1-1/\pi}{n} = n\left(-\dfrac{1/\pi}{n}\right) + n\left(\dfrac{1-1/\pi}{n}\right) = 1 - \dfrac{2}{\pi}$

As the number of subdivisions $2n$ gets larger, $R-L = 1 - \dfrac{2}{\pi}$ does not get smaller.

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Interesting counter example. Also, I have a more simple counter-example below. –  Garrett Jul 9 at 10:11

A non-rigorous answer: The definite integral, $$\int_a^bf(x) \ dx$$ gives the exact area between a function $f$, the $x$ axis, and the vertical lines $x=a$ and $x=b$. This integral is based upon the method of dividing the area under the curve into rectangles, and taking their total area. However, to get the exact answer, you have to have a huge amount of rectangles. One way of ensuring this is to let the width of each rectangle approach zero, which means the subintervals in the interval $[a,b$ become more numerous.

The more and more sub intervals, or the skinnier the rectangles, if you will, the more accurate the answer is regarding the area under the curve. We know that the left hand and the right hand give estimates that are larger and smaller than the true value. As the number of sub intervals increases, the more accurate each of these measures become, which means that the left and right hand estimates each approach the true value given by the integral, therefore the value between them must decrease.

Edit. It must be noted that this is usually the case, with such functions as polynomials, trigonometric, rational, and root functions. There are counter examples, such as those provided in other answers, that disallow this notion to be a rule.

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+1. Nice answer. –  Sanath Jul 9 at 2:19
    
I don't think this is correct. See my answer for a counter example. –  Garrett Jul 9 at 10:11
    
The question stated why does the difference between the two measures get smaller, not larger. In your example, they stay the same, which does not contradict my answer. –  J. Finnegan Jul 9 at 10:13
    
Ignore the first sentence of that comment. Also check your work, the left and right riemann sums are not always equal to 1 when the sub intervals become more numerous –  J. Finnegan Jul 9 at 13:29
    
The left-hand and right-hand estimates can give estimates that are smaller, the same or larger than the true value. It's even possible for both estimates to to be larger than (or in my counter-example, equal to) the true value. In other words, we know absolutely nothing about the relation between these estimates and the true value. –  Garrett Jul 9 at 19:58

A concise answer is as follows. As the number of subdivisions get larger, if the function is integrable, then the left-hand and the right-hand sums reach the integral of that function over the given interval; in other words, they become the same as the number of subdivisions goes to infinity.

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It is false that "the difference between the left-hand sum and right-hand sum always gets smaller as the number of subdivisions gets larger".

Here's a counter example. Take the function defined on the interval $[-1,1]$ by

$$f(x) = \begin{cases}-x+1 & x\in[0,1] \\ x+1 & x\in[-1,0] \end{cases}\;.$$

Consider breaking it into two equal subdivisions, $[-1,0]$ and $[0,1]$. Both the right-hand and left-hand riemann sums equal $1$ which is in fact the area under the curve.

Breaking it into four subdivisions, $[-1,-\frac{1}{2}, \frac{1}{2}, 1]$, both of the Riemann sums are again $1$, and therefore the difference between the right-hand and left-hand Riemann sums is still $0$.

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If you are only interested in going from $2$ subdivisions to $4$ (as opposed to an arbitrarily large number of subdivisions), then we can just consider $f(x) = 0$ on the interval $[0,1]$. –  JimmyK4542 Jul 9 at 20:13
    
Right, I guess $f(x)=0$ is the most simple counter-example. I think that should be an answer. –  Garrett Jul 9 at 20:32
    
As for showing just two examples of subdivisions, that's all I needed to show to disprove the statement that the difference between the left-hand and right-hand estimates always get smaller when the number of subdivisions get larger. If I wanted, I could have said that you should then subdivide into 6 equal subdivisions, 8 equal subdivisions, etc. but it would have been overkill. –  Garrett Jul 9 at 20:35

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