If $f$ is a measurable function, $g$ is an integrable function, and $|{f}| \leq |{g}|$ on $X$, prove that these imply $f$ is an integrable function and $$\int |{f}| d \mu \leq \int |{g}| d \mu.$$
$\textbf{Attempt:}$ I know I have to show that $\int |f| d \mu = \int f_{+} d \mu + \int f_{-} d \mu \leq \int |g| d\mu < \infty $ since $g$ is integrable and therefore $f$ is integrable. How can I show that $\int |f| d \mu $ is bounded by $\int |g| d\mu $ ?
$\int|g|d\mu-\int|f|d\mu=\int{(|g|-|f|)d\mu}=\int{|g-f|d\mu}$. The last equality is true because $g\geq f$.
As $|g-f|\geq 0$, we have $\int{|g-f|d\mu}\geq 0$. Hence, $\int|g|d\mu-\int|f|d\mu\geq 0$, or $$\int|g|d\mu\geq \int|f|d\mu$$