Why does the difference between the left-hand and right-hand sum get smaller with more subdivisions? 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.  
 A: 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.
A: 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.
A: 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.
A: 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 $0$.
