The following question is taken from Royden's $4$th Real Analysis, Chapter $4,$ question $16:$
Question $16:$ Let $f$ be a nonnegative bounded measurable function on a set of finite measure $E.$
Assume $\int_E f = 0.$
Show that $f=0$ a.e. on $E.$
My attempt:
Let
$$A :=\{ x\in E: f(x)=0 \}$$ and $$B:=\{x\in E: f(x)>0\}.$$
Then
$$\int_E f = \int_A f + \int_B f = \int_B f = 0.$$
Since $f$ is positive on $B,$ in order for $\int_B f=0,$ one must have $m(B)=0,$ that is, Lebesgue measure of $B$ must be zero.
Therefore, $f=0$ a.e.
I am not sure whether the bolded sentence is correct or not. If yes, can someone provide a proof. Otherwise, can someone provide an example to show the sentence is false?
Best Answer
Yes, it is correct, and it's the crucial part of this exercise. To flesh it out, instead of just considering $B = \{x: f(x)>0\},$ consider the countable partition $$B_1 = \{x:f(x)\ge 1\} \\B_2 = \{x: 1/2 \le f(x) < 1\}\\ B_3 = \{x: 1/3\le f(x) < 1/2\} $$ etc. Then we have $$ \int_{B}f(x) d\mu = \sum_{n=1}^\infty \int_{B_n} f(x)d\mu \ge \sum_{n=1}^\infty \mu(B_n) \frac{1}{n}.$$ The only way we can have $\int_B f(x) d\mu = 0$ is to have $\mu(B_n) = 0$ for all $n,$ which implies $\mu(B) = 0.$