ডারবো (Darboux) সমষ্টি কী?

One of the ways to approach the definite integral is by using Darboux sums. The upper Darboux sum will overestimate the definite integral, while the lower Darboux sum will underestimate the definite integral. As we refine these upper and lower Darboux sums, their values will (hopefully) become closer and closer to the actual value of the definite integral.

Definition: A partition $P$ of the interval $[a,b]$ is a set $\{x_0,\ldots,x_n\}$, such that $a=x_0<x_1<\cdots<x_n=b$.

Example: Some partitions of $[0,1]$ are $\{0,1\}$, $\left\{0,\frac{1}{2},1\right\}$, $\{0,0.01, 0.25, 0.4, 1\}$.

As their name suggests, partitions take an interval and break it up into various subintervals; these are the intervals $[x_{i-1}, x_{i}]$.

The idea with Darboux sums is that we want to take these subintervals and make them small enough that our function $f$, the function we want to integrate, is roughly constant on each subinterval. Then it is easy to estimate the value of $f$ on that subinterval, which will let us approximate the definite integral.

Definitions: The upper Darboux sum for a partition $P=\{x_0,\ldots,x_n\}$ on $[a,b]$ is defined as $$U(f,P):=\sum_{i=1}^nM_i\cdot(x_i-x_{i-1}),$$ where $M_i:=\sup_{[x_{i-1},x_i]}f(x)$, the supremum of $f$ on the subinterval $[x_{i-1},x_{i}]$.

Similarly, the lower Darboux sum for $P$ is $$L(f,P):=\sum_{i=1}^nm_i\cdot(x_i-x_{i-1}),$$ where $m_i:=\inf_{[x_{i-1},x_i]}f(x)$, the infimum of $f$ on the subinterval $[x_{i-1},x_{i}]$.

To get an intuitive idea of these Darboux sums, let's consider $f$ continuous. Then by the extreme value theorem, $f$ obtains its maximum on a compact interval like $[x_{i-1},x_i]$, and so we can replace the supremums with maximums.

The upper sum $U(f,P)$ takes the maximum $M_i$ of $f$ on each subinterval, then multiplies that value by the length of the subinterval. Graphically, what does this product $M_i\cdot(x_i-x_{i-1})$ represent?

Graphically, $M_i\cdot(x_i-x_{i-1})$ is the area of a rectangle - the one with base the subinterval $[x_{i-1},x_i]$, and height $M_i$.

Similarly, since $m_i$ is the minimum, the lower sum $L(f,P)$ will underestimate the value of the integral.

TODO: graph, picture of sums

Example: Let $f(x)=x^2$, and let $P=\{0,1,2\}$ be a partition of $[0,2]$. Let's compute $U(f,P)$. We need to compute $M_i$ on each interval. Since $f$ is increasing on $[0,2]$, the maximum on each subinterval will occur at the right-most point. Thus $$U(f,P)=\sum_{i=1}^2M_i(x_i-x_{i-1})=f(1)(1-0)+f(2)(2-1)=1+4=5$$

Now we have an upper bound, $U(f,P)$, and a lower bound, $L(f,P)$ for the integral. How can we improve our estimate?

By taking smaller and smaller subintervals $[x_{i-1},x_i]$, we can better approximate $f$ on each subinterval, which will give a better approximation for the integral.

TODO: refinements => U(f,P) decreases, L(f,P) increases => inf, sup in definition, why exist

Definitions: The upper Darboux integral $U(f)$ is defined as $$U(f):=\inf\{U(f,P):P\text{ partition of }[a,b]\}.$$ Similarly, the lower Darboux integral $L(f)$ is defined as $$L(f):=\sup\{L(f,P):P\text{ partition of }[a,b]\}.$$ If $U(f)=L(f)$, then we say $f$ is integrable on $[a,b]$, and write $$\int_a^bf=\int_a^bf(x)\,dx=U(f)=L(f).$$

Computing $U(f)$ and $L(f)$ is usually pretty difficult, but we'll look at one specific example where it's relatively easy.

Example (indicator function of the rationals, also called Dirichlet function). We're looking at the function $\chi_{\mathbb{Q}}$ defined by $$\chi_{\mathbb{Q}}(x):=\begin{cases}1,&x\in\mathbb{Q}\\0,&x\not\in\mathbb{Q}\end{cases}$$ Kind of a weird function, right? It's not continuous anywhere (try showing this! Use the fact that any interval contains both rational and irrational points, and so the function takes on values $0$ and $1$ in every $\delta-$interval. Can this function have a limit?)

Anyway, this is a classic example of a function that is not integrable. We'll show this by computing the upper and lower Darboux integrals and showing they differ. For convenience, we'll look at the interval $[0,1]$ and let $f$ denote the function $\chi_\mathbb{Q}$.

Let $P=\{0=x_0<x_1<\ldots<x_{n-1}<x_n=1\}$ be a partition of $[0,1]$. Note that any interval $[x_{i-1},x_i]$ contains both rational and irrational numbers, so that $f$ takes exactly the values $0$ and $1$. Thus $M_i=\sup_{[x_{i-1},x_i]}f(x)=1$, and $m_i=\inf_{[x_{i-1},x_i]}f(x)=0$.

Now we can compute $$U(f,P)=\sum_{i=1}^nM_i(x_{k}-x_{k-1})=1$$ and $$L(f,P)=\sum_{i=1}^nm_i(x_k-x_{k-1})=0.$$ These values for $U(f,P)$ and $L(f,P)$ hold for any partition $P$, and so we conclude $$U(f)=\inf_PU(f,P)=\inf_P1=1$$ and $$L(f)=\sup_PL(f,P)=\sup_P0=0.$$ Thus $U(f)\ne L(f)$, so $f$ is not integrable.

Usually it's pretty difficult and time-consuming to explicitly calculate $U(f)$ and $L(f)$.

Our main goal will be:

Theorem (Riemann's condition for integrability): A bounded function $f$ on $[a,b]$ is integrable if and only if for every $\varepsilon>0$, there exists a partition $P$ of $[a,b]$ such that $$U(f,P)-L(f,P)<\varepsilon.$$

TODO: Finish the rest