Uniform bounded of Riemann-like sum and improper integral For any $h>0$, suppose $\{(y_i,y_{i+1}]\mid i\in \mathbb{Z}\}$ be a uniform partition of $\mathbb{R}$ with mesh size $h$. I am considering under what condition for a continuous transition density function $p(h,x,y)$ of a stochastic process, we can have
$$\sum_i p(h,x,\xi)h^3\le Mh^\alpha$$ 
uniformly in $h,x$. Let's take Gaussian density function as an example.
That is, given the Gaussian density function
$$p(h,x,y)=\frac{1}{2\pi\sqrt{h}}e^{-\frac{(x-y)^2}{2h}},$$
I am considering the following two questions:


*

*Whether there exists $M,h_0>0$ independent of $x$, such that  $h\in (0,h_0)$, we have
$$\sum_i p(h,x,\xi_i)h^3\le Mh^\alpha,\tag{1}$$ for some $\alpha>0$. 

*Whether the Riemann-like sum can be approximated by a corresponding improper integral in the sense that for any $\epsilon>0$, there exists $h_0$ independent of $x$, such that for any $0<h<h_0,x\in \mathbb{R}$, we have
$$|\sum_i p(h,x,\xi_i)h- \int_\mathbb{R}p(h,x,y)\,dy|<\epsilon .\tag{2}$$


Here $\xi_i\in (y_i,y_{i+1})$ satisfies $p(h,x,\xi_i)=\sup_{y\in [y_i,y_{i+1}]}p(h,x,y)$.

I am really sorry that I edited the question. I added more background information so that I can express my original question in a clearer way. 
Thanks for Kirvich Entracus's answer. Generalizing the idea in the answer, if 
$$p(h,x,y)\le \dfrac{M}{\sqrt{h}}$$
uniformly and such that for fixed $h$, there exists $\delta(h)>0$, such that  for any fixed $x$, if $|y-x|>\delta(h)$, we have $p(h,x,y)$ decreases when $|y-x|$increases, 
then we can have 
$$\sum_i p(h,x,\xi_i)h\le \dfrac{M}{\sqrt{h}} \delta(h)+1$$
uniformly for small $h$ by comparing it with the improper integral.
 A: We may assume $y_k=kh$ $(k\in{\mathbb Z})$ and $0\leq x\leq h$. Then it is easy to see that
$$|x-x_k|\geq\bigl(|k|-1\bigr)h\qquad(k\ne0)\ .$$
It follows that
$$p(h,x,x_k)={1\over 2\pi\sqrt{h}}e^{-(x-x_k)^2/(2h)}\leq{1\over 2\pi\sqrt{h}}e^{-(|k|-1)^2h/2}\qquad(k\ne0)\ .$$
Summing over $k\in{\mathbb Z}$ and taking proper care of $k=0$ we therefore obtain
$$\sum_k p(h,x,x_k)\leq{1\over 2\pi\sqrt{h}}\bigl(2+\sum_k e^{-k^2 h/2}\bigr)={1\over 2\pi\sqrt{h}}\bigl(2+{1\over\sqrt{h}}\cdot\sqrt{h}\sum_k e^{-(k\sqrt{h})^2/2}\bigr)\ .$$
There is an $h_0>0$ such that
$$2\sqrt{h}<1,\qquad \bigl(\sqrt{2\pi}\sim\bigr)\  \sqrt{h}\sum_k e^{-(k\sqrt{h})^2/2}<3\qquad(0<h<h_0)\ .$$
It follows that
$$\sum_k p(h,x,x_k)\leq{2\over \pi h}\qquad(0<h<h_0)\ ,$$
so that you obtain the desired estimate with $\alpha=2$, $M={2\over\pi}$.
A: Looks not that complicated, if I understood the statements correctly.
For simplicity let me set $x=0$, $y_0=0$.
$i\geq 0\,\, \sum_{i\geq 0} p(h,0,\xi _i)h\leq \frac{h^\frac{1}{2}}{2\pi}+ \sum_{i\geq 0} q(h,0,\xi_i)h\leq\frac{h^\frac{1}{2}}{2\pi}+\int _{y\geq0}p(h,0,y)dy\leq \frac{h^\frac{1}{2}}{2\pi}+\frac{1}{2}\,\,\, where\,\, q(h,0,\xi_i)=inf\{p(h,0,y)|y\in (y_i,y_{i+1}]\}
\\
i<0\,\, \sum_{i< 0} p(h,0,\xi _i)h\leq \frac{h^\frac{1}{2}}{2\pi}+ \sum_{i< 0} q(h,0,\xi_i)h\leq\frac{h^\frac{1}{2}}{2\pi}+\int _{y<0}p(h,0,y)dy\leq\frac{h^\frac{1}{2}}{2\pi}+\frac{1}{2}\,\,\, where\,\, q(h,0,\xi_i)=inf\{p(h,0,y)|y\in (y_i,y_{i+1}]\}$
So for some constant $C>0$
$\sum_ip(h,0,y)h\leq Ch^{\frac{1}{2}}+1
\\
\sum_ip(h,0,y)h^3\leq (Ch^{\frac{1}{2}}+1)h^2$
So we can choose $(Ch_0^{\frac{1}{2}}+1)$ for M and $\alpha=2$.
When $x$ is not zero, slight modifications conclude the same result.
A: Suppose $f:\mathbb R \to [0,\infty)$ is even, decreasing on $[0,\infty),$ with $\int_{\mathbb R} f <\infty.$ Let $g$ be any translate of $f$ and let $c_i\in [(i-1)h,ih]$ for each $i\in \mathbb Z.$ Then
$$\tag 1 \int_{\mathbb R} g - 3f(0)h \le \sum_{i\in \mathbb Z} g(c_i)h \le \int_{\mathbb R} g + 3f(0)h.$$
This result answers your questions 1. and 2. in the affirmative: Your function $p(h,x,y),$ which we can think of as the $g$ in $(1),$ is the $x$ translate of $e^{-y^2/2h}/\sqrt {2\pi h},$ which we can think of at the $f$ above. The sum $\sum p(h,x,c_i)h$ is thus precisely the sum in $(1),$ hence is bounded above by $3/\sqrt {2\pi h} + 1$ for all $h>0.$ Thus the answer to 1. is yes, with $\alpha = 2$ (as others have noticed). For 2. we see
$$ |\sum_{i\in \mathbb Z} g(c_i)h - \int_{\mathbb R} g| \le 3f(0)h = 3\sqrt {h/2\pi}.$$
This gives a rate of vanishing on the order of $\sqrt h,$ so the answer to 2. is yes.
Proof of $(1)$: First observe that we have
$$\tag 2 2\sum_{k =1}^\infty f(kh)h \le \int_{\mathbb R} f \le 2\sum_{k =0}^\infty f(kh) h.$$
You prove this by first looking at $[0,\infty),$ where $f$ is decreasing, then do what you do when you prove the integral test of calculus. (It's just a matter of staring down some rectangles.) Then use the fact that $f$ is even; that's where the factor of $2$ comes from.
Fix $x\in \mathbb R$ and set $g(y) = f(y-x).$ Put $I_{k} =[(k-1)h, kh], k \in \mathbb Z.$ Now $x\in I_{k_0}$ for some $k_0.$ On $I_{k_0},$ $g\le f(0).$ On $I_{k_0+1}$ we also have $g\le f(0).$ On $I_{k_0+2}$ we have $g\le f(h),$ on $I_{k_0+3}$ we have $g\le f(2h),$ and so on down the line. The analogous estimates hold to the left of $I_{k_0}.$ Therefore
$$\sum_{i\in \mathbb Z} g(c_i)h  \le 3f(0)h + 2\sum_{k=1}^{\infty}f(kh)h \le 3f(0)h+ \int_{\mathbb R} f,$$
where we have used $(2).$ Because $\int_{\mathbb R} f = \int_{\mathbb R} g,$ we have the right hand inequality in $(1).$
For the left hand inequality in $(1),$ we have $g\ge f(h)$ on $I_{k_0},$ $g\ge f(2h)$ on $I_{k_0+1},$ $g\ge f(3h),$ and so on. The analogous estimates hold on the intervals to the left of $I_{k_0}.$ Therefore
$$ f(h)h + 2\sum_{k=2}^{\infty}f(kh)h \le \sum_{i\in \mathbb Z} g(c_i)h .$$
Now add $2f(0)h + f(h)h$ to both sides and use $(2).$ We get
$$\int_{\mathbb R} f  \le 2\sum_{k =0}^\infty f(kh)\cdot h \le 2f(0)h + f(h)h + \sum_{i\in \mathbb Z} g(c_i)h \le 3f(0)h + \sum_{i\in \mathbb Z} g(c_i)h.$$
This gives what we want, again using $\int_{\mathbb R} f = \int_{\mathbb R} g.$
