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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?

  • 0
    Take a slightly finer partition that encloses the line $\{\frac{1}{2}\} \times [0,1]$ in a rectangle of area < \epsilon. Then it should be straightforward? The only issue is this line, which has content 0.2012-07-26

2 Answers 2

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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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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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    @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 $[1/2,1]\times[0,1]$.2012-07-27