[proofplan] The proof compares $X$ with its pointwise absolute value. Since $-|X| \le X \le |X|$ on $\Omega$, positivity and linearity of the [Lebesgue integral](/page/Lebesgue%20Integral) give two scalar inequalities, namely $\mathbb E[X] \le \mathbb E[|X|]$ and $-\mathbb E[X] \le \mathbb E[|X|]$. These two inequalities are exactly the statement that the absolute value of $\mathbb E[X]$ is bounded by $\mathbb E[|X|]$. [/proofplan]

[step:Use integrability to make all expectations finite] Since $X \in L^1(\Omega,\mathcal F,\mathbb P)$, the [random variable](/page/Random%20Variable) $|X|:\Omega\to\mathbb R$ is integrable. Therefore both \begin{align*} \mathbb E[X]=\int_\Omega X\,d\mathbb P(\omega) \end{align*} and \begin{align*} \mathbb E[|X|]=\int_\Omega |X|\,d\mathbb P(\omega) \end{align*} are finite [real numbers](/page/Real%20Numbers). [/step]

[step:Integrate the pointwise upper bound $X\le |X|$]Define the non-negative integrable random variable \begin{align*} Y:\Omega\to\mathbb R \end{align*} by \begin{align*} Y(\omega)=|X(\omega)|-X(\omega) \end{align*} for $\omega\in\Omega$. Since $Y(\omega)\ge 0$ for every $\omega\in\Omega$, positivity of the Lebesgue integral gives \begin{align*} 0\le \int_\Omega Y\,d\mathbb P. \end{align*} By linearity of the Lebesgue integral for integrable functions, \begin{align*} 0\le \int_\Omega |X|\,d\mathbb P-\int_\Omega X\,d\mathbb P. \end{align*} Thus \begin{align*} \mathbb E[X]\le \mathbb E[|X|]. \end{align*}[/step]

[guided]The first comparison comes from the pointwise inequality $X\le |X|$. To turn that pointwise inequality into an inequality between expectations, we subtract the left-hand side from the right-hand side and integrate a non-negative function. Define \begin{align*} Y:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R)) \end{align*} by \begin{align*} Y(\omega)=|X(\omega)|-X(\omega) \end{align*} for $\omega\in\Omega$. The function $Y$ is measurable because $X$ is measurable and the absolute value map on $\mathbb R$ is Borel measurable. Also $Y$ is integrable because \begin{align*} |Y(\omega)|=||X(\omega)|-X(\omega)|\le 2|X(\omega)| \end{align*} for every $\omega\in\Omega$, and $|X|$ is integrable by the hypothesis $X\in L^1(\Omega,\mathcal F,\mathbb P)$. For every $\omega\in\Omega$, the real number $|X(\omega)|-X(\omega)$ is non-negative. Hence positivity of the Lebesgue integral gives \begin{align*} 0\le \int_\Omega Y\,d\mathbb P(\omega). \end{align*} Substituting the definition of $Y$ and using linearity of the Lebesgue integral for integrable functions yields \begin{align*} 0\le \int_\Omega |X|\,d\mathbb P(\omega)-\int_\Omega X\,d\mathbb P(\omega). \end{align*} By the definition of expectation, this is \begin{align*} 0\le \mathbb E[|X|]-\mathbb E[X]. \end{align*} Therefore \begin{align*} \mathbb E[X]\le \mathbb E[|X|]. \end{align*}[/guided]

[step:Integrate the pointwise lower bound $-|X|\le X$] Define the non-negative integrable random variable \begin{align*} Z:\Omega\to\mathbb R \end{align*} by \begin{align*} Z(\omega)=|X(\omega)|+X(\omega) \end{align*} for $\omega\in\Omega$. Since $Z(\omega)\ge 0$ for every $\omega\in\Omega$, positivity of the Lebesgue integral gives \begin{align*} 0\le \int_\Omega Z\,d\mathbb P(\omega). \end{align*} By linearity of the Lebesgue integral for integrable functions, \begin{align*} 0\le \int_\Omega |X|\,d\mathbb P(\omega)+\int_\Omega X\,d\mathbb P(\omega). \end{align*} Thus \begin{align*} -\mathbb E[X]\le \mathbb E[|X|]. \end{align*} [/step]

[step:Combine the two scalar inequalities] From the preceding two steps, \begin{align*} \mathbb E[X]\le \mathbb E[|X|] \end{align*} and \begin{align*} -\mathbb E[X]\le \mathbb E[|X|]. \end{align*} For a real number $a$, the inequalities $a\le b$ and $-a\le b$ imply $|a|\le b$. Applying this with $a=\mathbb E[X]$ and $b=\mathbb E[|X|]$ gives \begin{align*} |\mathbb E[X]|\le \mathbb E[|X|]. \end{align*} This is the desired bound. [/step]