Define $B:=\big\{(x,y)\in [0,1]^2:xy\leq 1/2\big\}$.
Let $\lambda ^2$ be the Lebesgue measure in $\mathbb{R}^2$. How can I calculate $\lambda ^2(B)$?
I tried to obtain that value calculating the Lebesgue integral of the indicator function $\mathbf{1}_B$, but I couldn't find a nice way to express this function in order to calculate the integral.
Please help me!
Define $B:=\{(x,y)\in[0,1]^2:xy\leq1/2\}$. From Fubini's Theorem, $$ \text{Area of B} = \int_{[0,1]^2}1_B dxdy = \int_0^1\left(\int_0^{\min\{1,\frac1{2x}\}}1 dy \right)dx. $$ So the area of $B$ is given by $$ \int_0^1 \min\left\{1,\frac{1}{2x}\right\} dx. $$ In other words, the area in the square, below the curve given by $\dfrac1{2x}$. The $\min\{\cdot,\cdot\}$ makes sure you do not go above the top side of the square. Hence, the area is $$ \int_0^{1/2} 1 dx + \int_{1/2}^1 \frac{1}{2x} dx = \frac{1}{2} + \frac{1}{2}\left[ \ln x\right]_{1/2}^1 = \frac{1}{2}\left(1 + 0 - \ln \left( \frac{1}{2} \right) \right)= \frac{1}{2}(1 + \ln 2). $$
Update: Thanks to @Mark below who spotted my error. I've updated the answer.