Let $(\Omega, \mathscr A, \lambda)$ be a measure space, $a \ge 0$ a constant and $f: \Omega \rightarrow [0, \infty]$ be a measurable function. Show that
$a * \lambda(f \ge a) \le \int_{\Omega} f(x) d\lambda(x)$.
I thought it might be easier to prove this thing while assuming that $\lambda$ is the Lebesgue-measure first, and I would like to know whether my proof is correct or not.
Let $\Omega = \Bbb R$ and $\lambda$ be the Lebesgue-measure.
Since $f$ is measurable and non-negative, we know that there is a sequence of increasing simple functions $f_k$ with $f_k \uparrow f$ such that
$\int_{\Omega} f d\lambda = \lim_{k \rightarrow \infty} \int_{\Omega} f_k$.
Furthermore, we are allowed to assume that a $\le \sup(f)$, because for $a > \sup(f)$, the set $(f \ge a)$ would be empty (since there can't be no such $x \in \Omega)$ and the inequality would be true trivially. We write:
$a * \lambda(f \ge a) = a * c$ for some constant $c \in \Bbb R_+$, which is just the length of the interval of $(f \ge a)$. We define
$g := a * c$,
which is just the volumn of the rectangle spanned by $a$ and $c$. Now, since $a \le \sup(f)$, we know that there must be a $k_0 \in \Bbb N$ such that $g \le f_k$ for every $k \ge k_0$, and since $f_k \uparrow f$, we get that $g \le f$, and hence
$g = \int_{(f \ge a)} a * x$ $d\lambda(x) \le \int_{\Omega} f(x)$ $d\lambda(x)$.
Much too complicated…
$$\int_{\Omega} f d\lambda = \int_{\{f<a\}} f d\lambda + \int_{\{f\ge a\}} f d\lambda \ge 0 + \int_{\{f\ge a\}} a d\lambda = a\lambda\left(\{f\ge a\}\right)$$