# Show that the function $f: [0, 1] \times [0, 1] \to \mathbb{R}$ is integrable

Let $f: [0, 1] \times [0, 1] \to \mathbb{R}$ be defined by

$$f(x, y) = \begin{cases} 0,&\text{if } 0 \le x < \frac{1}{2}\\ 1,&\text{if }\frac{1}{2} \le x \le 1\;. \end{cases}$$

I need to show that this function is integrable, and my instructor says if we consider the partition $P = (P_1, P_2)$ where $P_1 = P_2 = \{0, \frac{1}{2}, 1\}$, then $U(f, P) = L(f,P) = \frac{1}{2}$, but this is clearly false since

$$U(f, P) = (1/2)^2 + (1/2)^2 + (1/2)^2 + (1/2)^2 = 1$$

and

$$L(f, P) = 0 + 0 + (1/2)^2 + (1/2)^2 = 1/2.$$

Can someone point out where I am going wrong?

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We don't even have a partition the way it is defined – Jean-Sébastien Jul 26 '12 at 22:14
I think $P$ is a partition of $[0, 1] \times [0, 1]$ – WacDonald's Jul 26 '12 at 22:15
Jean-Sébastien is right. Specifying a grid is not a description of partition – Norbert Jul 26 '12 at 22:15
I understand that your partition is supposed to be 4 squares, but you need explcitly and very carefully describe all this 4 squares. Some of them will contain their boundary, others doesn't – Norbert Jul 26 '12 at 22:17
I interpret what I see as cutting into four equal squares. – ncmathsadist Jul 26 '12 at 22:18

## 2 Answers

Take $$P_1=[0,\frac{1}{2})\times[0,1], P_2=[\frac{1}{2},1]\times[0,1].$$ Then $$U(f,P)=|P_1|\sup_{P_1}f+|P_2|\sup_{P_2}f=|P_2|=\frac{1}{2}, \ L(f,P)=|P_1|\inf_{P_1}f+|P_2|\inf_{P_2}f=|P_2|=\frac{1}{2}.$$

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In my question $P_i$ are partitions of the interval and not subrectangles, but I see what you did. So you can just throw out the endpoint and call $[0, \frac{1}{2} ) \times [0, 1]$ a subrectangle? – WacDonald's Jul 26 '12 at 23:36
@WacDonald's I don't know which definition you're using, but your function is clearly a simple function, in fact it is the characteristic function of $[1/2,1]\times[0,1]$. – Mercy King Jul 27 '12 at 9:10

Let's try the simpler one-dimensional integral $\int_0^1 f(x) \, dx$ with piecewise defined function:

$$f(x) = \left\{ \begin{array}{cc} 1 & x > 1/2 \\ \\ 0 & x < 1/2 \end{array} \right.$$

Use the partition $[0, 1/2 - \epsilon]\cup [1/2 - \epsilon, 1/2 + \epsilon] \cup [1/2 - \epsilon, 1]$. Then

$U(f,P) = 0\cdot (1/2 - \epsilon) + 2\epsilon\cdot1 + 1 \cdot(1/2 - \epsilon)= 1/2 + \epsilon$ $L(f,P) = 0\cdot (1/2 - \epsilon) + 2\epsilon\cdot0 + 1 \cdot(1/2 - \epsilon)= 1/2 - \epsilon$

Then we have upper and lower bound for the integral: $1/2 - \epsilon < \int_0^1 f(x) dx < 1/2 + \epsilon$

This can be extended to 2 dimensions.

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