Let $f$ be measurable and $a,b\in\mathbb{R}$ with $\frac{1}{\lambda(M)}\int_Mf\ d\lambda \in [a,b]$ Show that: $f(x) \in [a,b]$ almost everywhere.

Question

Assignment:

Let $f$ be Lebesgue - measurable and $a,b \in \mathbb{R}$ with the property: $$\frac{1}{\lambda(M)} \cdot \int_Mf\ d\lambda \in [a,b]$$ for all Lebesgue - measurable sets $M \subset \mathbb{R}^n$ with $0 < \lambda(M) < \infty$.

Show that: $f(x) \in [a,b]$ almost everywhere.

So I need to show that the set $N$ defined as $N:= \{x\in\mathbb{R}^n; \ f(x) \notin [a,b]\}$ has a measure of zero. However, I cannot really see how to go from there and how the property given could help me.

I'd appreciate any help.

Answer 1

Suppose $N = \{x:f(x)\notin[a,b]\}$ is not a null set.

Then at least one of $N_1 = \{x:f(x)<a\}$ or $N_2 = \{x:f(x)>b\}$ is not a null set since $N = N_1\cup N_2$

If $N_1$ is not a null set, then (since $f(x)<a$ on $N_1$)

$$\frac{1}{\lambda(N_1)}\int_{N_1}f~d\lambda<\frac{1}{\lambda(N_1)}\int_{N_1}a~d\lambda=a$$

If $N_2$ is not a null set, then(since $f(x)>b$ on $N_2$)

$$\frac{1}{\lambda(N_2)}\int_{N_2}f~d\lambda>\frac{1}{\lambda(N_2)}\int_{N_2}b~d\lambda=b$$

Either way you contradict your assumption.

Answer 2

Since $[a,b]^{c}$ is open we can write it as a countable union of disjoint intervals. So it is enough to prove that $\lambda(f^{-1}(I))=0$, where $I:={B(x,r)}\subseteq [a,b]^{c}$. Write $E:=f^{-1}(I)$.

If $\lambda(E)>0$, then

$$\left|\frac{1}{\lambda(E)}\int_{E}f d\lambda-x\right|\leq\frac{1}{\lambda(E)}\int_{E}|f-x| d\lambda\leq r$$

a contradiction, since $\frac{1}{\lambda(E)}\int_{E}f d\lambda \in [a,b]$.