Consider $\mathbb{R}$ with the lesbesgue measure, show that there does not exist an $f$.

Question

Consider $\mathbb{R}$ with the lesbesgue measure, show that there does not exist an $f$.

with $\int_{\mathbb{R}}| f |< \infty$ and $\int_{\mathbb{R}}| f-1 |< \infty$

I have no idea where to start actually, the hint tells me that i need to show that functions with this property have:

$\lambda(x:f(x) \geq \frac{1}{2})$ and $\lambda(x:f(x) \leq \frac{1}{2})$ are finite

Any hints would be much appreciated!

Kees

Answer 1

If such an $f$ existed, then set $A = \{x : f(x) \ge 1/2\}$ and $B = \{x : f(x) \le 1/2\}$. So $$\lambda(A) = \int_{\Bbb R} 1_A = 2\int_{\Bbb R}\frac{1}{2}1_A \le 2\int_{\Bbb R} f1_A \le 2 \int_{\Bbb R} |f| < \infty$$ and $$\lambda(B) = \int_{\Bbb R} 1_B = 2\int_{\Bbb R} \frac{1}{2}1_B \le 2 \int_{\Bbb R} (1 - f)1_B \le \int_{\Bbb R} |1 - f| < \infty.$$ Then since $\Bbb R$ is the union of the $A$ and $B$, $\lambda(\Bbb R) \le \lambda(A) + \lambda(B) < \infty$, a contradiction.