Suppose $f$ be a non-negative function in $L^{1}[0,1]$. Suppose that for every integer $n=1,2,3..$, we have
$$ \int_{[0,1]} f(x)^{n} dm= \int_{[0,1]} f(x) dm $$
Then prove that $f(x)$ must be almost everywhere equal to the characteristic function $\chi_{E}$ of a measurable set.
My attempt:
Since $f \in L^{1}[0,1], \int_{[0,1]} f dm < \infty$. I am partitioning the sample space such into
$$A_{1} = \{ |f| > 1 \}$$
$$A_{2} = \{ |f| = 1 \}$$
$$A_{3} = \{ 0 < |f| < 1 \}$$
Applying Fatou's lemma to $g_{n}=f(x)^{n}$, we get that
$$\int_{[0,1]} \liminf_{n \to \infty} g_{n} \leq \liminf_{n \to \infty} \int g_n= \int_{[0,1]}f(x) dm $$
$$\int_{[0,1]} \liminf_{n \to \infty} f(x)^n \leq \int_{[0,1]}f(x) dm $$
Now, I multiply both the sides with $\chi_{A_1}(x)$ ( it is measurable as f is measurable).
$$\int_{[0,1]} \liminf_{n \to \infty} f(x)^n \chi_{A_1}(x) \leq \int_{[0,1]}f(x) \chi_{A_1}(x) dm $$
$$\int_{[0,1] \cap A_1} \liminf_{n \to \infty} f(x)^n \leq \int_{[0,1]\cap A_1}f(x) dm $$
Thus, $m(A_1)=0$, else we get a contradiction.
For $A_{3}$, I multiply the original equation with $\chi_{A_3}$ on both the sides,
$$ \int_{[0,1] \cap A_3} f(x)^{n} dm= \int_{[0,1]\cap A_3} f(x) > \int_{[0,1] \cap A_3} dm > 0$$
But, this is true for all $n$ and as we increase $n$ to infinity, we get $0$ on the Left Hand side, hence $m(A_3)=0$ as well.
For values of $x$ in $A_2$, $f(x)^{n}=f(x)$ for all $n$. Hence $f(x)=1$ almost everywhere.
The measurable set $E= \{|f|=1\}$.
Can someone verify or give alternative ideas?
Best Answer
As you have correctly identified, it suffices to show that $m(A_{1})= m(A_{3}) = 0$. We have $$\int f^n\chi_{A_1}+\int f^n\chi_{A_3} = \int f\chi_{A_1}+\int f\chi_{A_3}$$ for all $n$. Since $f^n\chi_{A_1}\nearrow \infty\cdot\chi_{A_1}$ and $f^n\chi_{A_3}\rightarrow 0$ as $n\rightarrow\infty$, we can use the Monotone Convergence Theorem for the first integral and the Dominated Convergence Theorem for the second integral to obtain $$\infty\cdot m(A_{1}) = \int f\chi_{A_1}+\int f\chi_{A_3}.$$ Since $f$ is integrable, we must have $m(A_{1}) = 0$ and the equation reduces to $$\int f\chi_{A_3} = 0.$$ Therefore, $f\chi_{A_3} = 0$ a.e. which gives $m(A_3) = 0$ (since $f\chi_{A_3} > 0$ on $A_{3}$).