Show that $h \circ f$ is measurable and that $\lambda(f \ge a) \le$ $1 \over h(a)$ $\int_{\Omega} (h \circ f) (x) \,d\lambda(x)$

Question

Let $(\Omega, \mathscr A, \lambda)$ be a measure space, $a > 0$ a constant and $f: \Omega \rightarrow \Bbb R$ a measurable function and $h: \Bbb R \rightarrow (0, \infty)$ be a monotonically increasing function.

Show that $h \circ f$ is measurable and that $\lambda(f \ge a) \le {1 \over h(a)} \int_\Omega (h \circ f) (x) \, d\lambda(x)$.

Well, I know that $h \circ f$ is measurable due to the fact that $h$ is monotonically increasing, hence measurable, and that every composition of measurable functions is measurable too. Furthermore, I know that

$\displaystyle\lambda(f \ge a) = \int_\Omega \chi_{(f \ge a)} \, d \lambda$, so we would have to show that

$$\int_\Omega \chi_{(f \ge a)} \le{1 \over h(a)}\int_\Omega h \circ f \, d\lambda.$$

I don't see what to do next though. It might be helpful to note that

$${1 \over h(a)}\int_\Omega h \circ f \, d\lambda = {1 \over h(a)} \left(\int_{(h \circ f < a)} h \circ f \, d\lambda+{1 \over h(a)} \int_{(h \circ f \ge a)} h \circ f \, d\lambda\right).$$

Answer 1

In the set $\{f \ge a\}$ we have $h\circ f \ge h(a)$ by monotonicity of $h$. Therefore, $$h(a) \lambda\{f \ge a\} = h(a) \int_\Omega \chi_{\{f \ge a\}}\,d\lambda = \int_{\{f \ge a\}} h(a) \, d\lambda \le \int_{\{f \ge a\}} h\circ f\,d\lambda \le \int_\Omega h\circ f\,d\lambda$$ and the result follows on dividing both sides by $h(a)$ (which is positive, hence not zero).

Answer 2

Since $h$ is increasing and positive, it follows that $$ h(a)1_{f(x)\geq a}\leq h(f(x)) $$ for all $x\in\Omega$. Integrating both sides of this inequality then leads to $$h(a)\lambda(f\geq a)\leq \int_{\Omega}h(f(x))\;d\lambda(x)$$