Suppose that $m(\{|f| > \lambda\}) \to 0$ whenever $\lambda \to \infty$ and let $$f^{\ast}(t) = \inf\{\lambda \mid m(\{|f| > \lambda\}) \leq t\},\quad t \geq0.$$ Prove that
(i) $f$ is nonincreasing and right-continuous.
(ii) If $0<\eta<m(\{|f|>\lambda\}) < \infty$, then $$f^{\ast}(m(\{|f|>\lambda\})) \leq \lambda < f^{\ast}(m(\{|f|>\lambda\}) - \eta).$$
(iii) If $f^{\ast}$ is continuous at $t = m(\{|f|>\lambda\})$, then $f^{\ast}(m(\{|f|>\lambda\})) = \lambda$.
(iv) $f$ and $f^{\ast}$ are equimeasurable.
- $m$ is the Lebesgue measure.
I proved the items (i) and (iv). In the item (ii), I proved the first inequality. I need help with the second inequality of item (ii) and item (iii). Can someone help me?
Denote $S_{\lambda}:=\{|f|>\lambda\}$.
(ii):
$$f^*(m(S_{\lambda}))) = \mbox{inf}\{\tau|\ m(S_{\tau})\leq m(S_{\lambda})\}\leq\lambda.$$
$$f(m(S_{\lambda})-\eta)=\mbox{inf}\{\tau|\ m(S_{\tau})\leq m(S_{\lambda})-\eta\},$$
Clearly the minimizer $\tau^*= \mbox{inf}\{\tau|m(S_\tau)\leq m(S_\lambda)-\eta\}$ above must satisfy $\tau^*>\lambda$, as $m(S_\tau)$ is monotonically decreasing in $\tau$, e.g. $m(S_\tau)\geq m(S_\lambda)$ whenever $\tau<\lambda$. .
(iii):
$$f^*(t)=f^*(m(S_{\lambda}))\leq \lambda,$$
by (ii). On the other hand for $\eta$ satisfying (ii),
$$\lambda<f^*(m(S_{\lambda})-\eta)$$
So take the limit of $\eta\rightarrow 0^+$. It follows that $\lambda\leq f(m(S_{\lambda}))$, so $\lambda \leq f(t) \leq \lambda$, i.e. $f(t)=\lambda$.