Let \(f: \RR^n\to \RR\) be a Lebesgue measurable function.
Prove that for all \(\alpha> 0\) , \[ A_ \alpha \da \ts{x\in \RR^n \st \abs{ f(x) } > \alpha} \implies m(A_ \alpha) \leq {1\over \alpha} \int \abs{f (x)} \dx .\]
Prove that \[ \int \abs{ f(x) } \dx = 0 \iff f = 0 \text{ almost everywhere} .\]
Solution
1. For \(\alpha > 0\), \(m(A_\alpha) \le \frac{1}{\alpha}\int|f|\).
Since \(|f| > \alpha\) on \(A_\alpha\) and \(|f| \geq 0\) everywhere, \(\int_{\RR^n}|f| \ge \int_{A_\alpha}|f| \ge \int_{A_\alpha}\alpha = \alpha\, m(A_\alpha)\).
2. \(\int |f| = 0\) if and only if \(f = 0\) a.e.
2.1. If \(f = 0\) a.e., then \(\int|f| = 0\).
\(|f|\) agrees a.e. with the zero function, and functions equal a.e. have equal integrals.
2.2. If \(\int|f| = 0\), then \(f = 0\) a.e.
\(\theset{|f| > 0} = \bigcup_{k \geq 1} A_{1/k}\), and step 1 gives \(m(A_{1/k}) \le k\int|f| = 0\) for each \(k\). A countable union of null sets is null.
2.3. Q.E.D.
Steps 1 and 2.
3. Q.E.D.
Step 1 is part (a) and step 2 is part (b).