Evaluate the double integral $\iint_{Q} f(x,y)\,dx\,dy$

Question

Let $f$ be defined on the rectangle $Q =[0,1] \times [0,1]$ as follows $$f(x,y) = \begin{cases} 1 \ \text{if}\ x=y \\0 \text{ if } x \neq y \end{cases}$$

Evaluate the double integral $\iint_{Q} f(x,y)\,dx\,dy$.

My attempt : First integrate with respect to $x$

$$\int_{0}^{1} \int_{0}^{1} f(x,y)\,dx\,dy=\int_{0}^{1}1 \,dy=1 $$

Is it true ?

Answer 1

You made a mistake in your substitution. You should have

$$\int_{0}^{1} \int_{0}^{1} f(x,y)\,dx\,dy =\int_{0}^{1} \left(\int_{0}^{1} 1_{\{y\}}(x)\,dx\right)\,dy =\int_{0}^{1}0 \,dy=0\;. $$

This is because the definition $f$ tells you that for every fixed $y\in[0,1]$, $$ f(x,y)=1_{\{y\}}(x),\quad x\in[0,1]\;, $$ where $1_A$ denotes the indicator function $$ 1_A(x)=\begin{cases} 1,& x\in A\;;\\ 0,& x\not\in A\;. \end{cases} $$

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Remark.

If you know what Lebesgue measure means as Filippo's answer points out, you can immediately have the value zero for the integral. Alternatively, you can refer to the definition of double integrals.

Answer 2

Hint:

$\int_{0}^{1} \int_{0}^{1} f(x,y)dxdy=0$

Just observe that the measure of $\{(x,y) \in \mathbb{R}^2: x=y\}$ has Lebesgue measure $0$.