Question integral

Question

Assume that $ \lambda $ is the Lebesgue measure on $ \mathbb{R}^N,\ N \geq 1. $ Let $ f \in L^1_\text{loc}( \mathbb{R}^N, d \lambda),\ f \geq 0. $ Assume that there exists $ \alpha > 0 $ such that, for all Borel set $ E $ of $ \mathbb{R}^N $ with $ \lambda(E) = \alpha, $ then $ \int_E f \, d \lambda = \alpha. $

The question is : prove that $ f = 1 \text{ a.e.} $

Any idea? (I have the idea to set $ D = \{x , f(x) < 1\} $ and to prove by contradiction that $ \lambda (D) = 0 $ and later the set $ D' = \{x , f(x) > 1\} $)

Answer 1

Your idea looks good to me. Both $D$ and $D'$ have to have measure less than $\alpha$, otherwise we could integrate over a subset of one with measure $\alpha$ and we'd be in trouble.

However, since both have measure less than $\alpha$, we can find Borel sets $E$ and $E'$ that are disjoint from both with $\lambda(E \cup D) = \alpha = \lambda(E' \cup D')$. Integrating over $E \cup D$ and $E' \cup D'$ is going to cause problems unless $D$ and $D'$ have measure $0$.

Answer 2

Suppose $\lambda(D) = \lambda \{x:f(x)<1\}>0.$

Let $D_n = \lambda \{x: 1/(n-1) \le f(x)< 1 - 1/n\}$ for $n=2,3,4,\ldots$ and $D_1= \{x:f(x)<1/2\}.$

By countable additivity, for some $n,$ $\lambda(D_n)>0.$

Think about $\displaystyle \int_{D_n} f\,d\lambda.$