If $X$ is an integrable random variable, then $\left|\int_{A}X\mathrm{d}\mathbb{P}\right| \leq \int_{A}|X|\mathrm{d}\mathbb{P}$ Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. If $X$ is an integrable random variable, then
\begin{align*}
\left|\int_{A}X\mathrm{d}\mathbb{P}\right| \leq \int_{A}|X|\mathrm{d}\mathbb{P}
\end{align*}
Here is my attempt:
\begin{align*}
\left|\int_{A}X\mathrm{d}\mathbb{P}\right| & = \left|\int_{A}(X^{+} - X^{-})\mathrm{d}\mathbb{P}\right|\\\\
& = \left|\int_{A}X^{+}\mathrm{d}\mathbb{P} - \int_{A}X^{-}\mathrm{d}\mathbb{P}\right|\\\\
& \leq \int_{A}X^{+}\mathrm{d}\mathbb{P} + \int_{A}X^{-}\mathrm{d}\mathbb{P}\\\\
& = \int_{A}(X^{+} + X^{-})\mathrm{d}\mathbb{P}\\\\
& = \int_{A}|X|\mathrm{d}\mathbb{P}
\end{align*}
Can someone criticize my solution?

The solution is correct. Alternatively, you can argue by noting that the left-hand side of the inequality is $c\int_A X\, d\mathbb{P}$ for some complex number $c$ with $|c| = 1$. Then write the integral as $\operatorname{Re}(c\int_A X\, d\mathbb{P}) = \int_A \operatorname{Re}(cX)\, d\mathbb{P}$ and observe $\operatorname{Re}(cX) \le |cX| = |X|$, ending the proof.
