$f$ is $μ$-integrable , then given $\epsilon > 0$, $ \exists δ > 0$ such that $\forall D ∈ A$ with $μ(D) < δ$, one has $\int_{D}|f|dμ < \epsilon $ 
Let $f$ be an extended-real valued measurable function
  on a measure space $(Ω, A, μ)$ . Show that if f is $μ$-integrable , then given $\epsilon > 0$, $ \exists δ > 0$ such that $\forall D ∈ A$ with $μ(D) < δ$, one has $\int_{D}|f|dμ < \epsilon $

Since, $\int_{\Omega} f < \infty \implies \int_{\Omega} |f| < \infty \implies \exists M > 0 : \int_{\Omega} |f| = M $
Hence for any set $D \in A$, $\int_{D} |f| d\mu=\int_{\Omega}|f|1_{D} d\mu $
How to complete the proof? I thought of Cauchy Schwarz but not sure whether :  $\int_{\Omega} |f| < \infty \implies \int_{\Omega} |f|^{2} < \infty $ 
Thanks in advance for help!
 A: This is a possible solution. We have that $f\in\mathcal L_1(\Omega,\mu,\overline{\Bbb R})$ (that is, $f$ is $\mu$-integrable), then $\|f\|_1:=\int_\Omega|f|\,d\mu<\infty$.
Because the set of $\mu$-simple functions $\mathcal S(\Omega,\mu,\Bbb R)$ is dense in $\mathcal L_1(\Omega,\mu,\overline{\Bbb R})$ then there is some $\varphi\in\mathcal S(\Omega,\mu,\Bbb R)$ such that $\|f-\varphi\|_1<\epsilon/2$ for any chosen $\epsilon>0$, then
$$
\int_A|f-\varphi|\,d\mu\le\int_A|f-\varphi|\,d\mu+\int_{A^\complement}|f-\varphi|\,d\mu=\|f-\varphi\|_1<\frac\epsilon2
$$
Hence
$$
\begin{align}\int_A |f|\,d\mu&\le\int_A|f-\varphi+\varphi|\,d\mu\\
&\le\int_A|f-\varphi|\,d\mu+\int_A|\varphi|\,d\mu\\&<\frac\epsilon2+\mu(A)\|\varphi\|_\infty\end{align}
$$
Can you follow from here? The solution in spoilers.

 Then for $\mu(A)<\frac{\epsilon}{2\|\varphi\|_\infty}$ we find that $\int_A |f|\,d\mu<\epsilon$, as desired.

P.D.: also, instead of the set of $\mu$-simple functions you can use the set of continuous functions with compact support, denoted by $C_c(\Omega,\Bbb R)$, what is also dense in $\mathcal L_1(\Omega,\mu,\overline{\Bbb R})$.
