[proofplan] We first check that all quantities in the formula are finite: the finite second moment implies that $X$ is integrable, and then the centered square is integrable. After setting $m:=\mathbb E[X]$, we expand $(X-m)^2$ pointwise and integrate the resulting identity. The probability normalization $\mathbb P(\Omega)=1$ then reduces the expression to $\mathbb E[X^2]-m^2$. [/proofplan] [step:Show that the expectation and variance are finite] Define $m\in\mathbb R$ by \begin{align*} m:=\mathbb E[X]=\int_\Omega X(\omega)\,d\mathbb P(\omega), \end{align*} after verifying that this integral is finite. For every $\omega\in\Omega$, \begin{align*} |X(\omega)|\leq \frac{X(\omega)^2+1}{2}, \end{align*} because $(|X(\omega)|-1)^2\geq 0$. Since $\mathbb P(\Omega)=1$ and $\mathbb E[X^2]<\infty$, monotonicity of the [Lebesgue integral](/page/Lebesgue%20Integral) gives \begin{align*} \int_\Omega |X(\omega)|\,d\mathbb P(\omega) \leq \frac{1}{2}\int_\Omega X(\omega)^2\,d\mathbb P(\omega) + \frac{1}{2}\int_\Omega 1\,d\mathbb P(\omega) <\infty. \end{align*} Thus $X$ is integrable and $m$ is well-defined. Now define the centered square [random variable](/page/Random%20Variable) \begin{align*} Y:(\Omega,\mathcal F)&\to(\mathbb R,\mathcal B(\mathbb R))\\ \omega&\mapsto (X(\omega)-m)^2. \end{align*} For every $\omega\in\Omega$, \begin{align*} Y(\omega)=(X(\omega)-m)^2\leq 2X(\omega)^2+2m^2. \end{align*} The right-hand side is integrable with respect to $\mathbb P$, so \begin{align*} \int_\Omega Y(\omega)\,d\mathbb P(\omega)<\infty. \end{align*} Therefore $\operatorname{Var}(X)=\mathbb E[Y]$ is finite. [/step] [step:Expand the centered square and integrate term by term] For every $\omega\in\Omega$, the algebraic identity for the real number $X(\omega)$ gives \begin{align*} (X(\omega)-m)^2=X(\omega)^2-2mX(\omega)+m^2. \end{align*} Each term on the right-hand side is integrable: $X^2$ is integrable by hypothesis, $X$ is integrable by the previous step, and the constant function $\omega\mapsto m^2$ is integrable because $\mathbb P(\Omega)=1$. Hence linearity of the Lebesgue integral yields \begin{align*} \operatorname{Var}(X) &=\int_\Omega (X(\omega)-m)^2\,d\mathbb P(\omega)\\ &=\int_\Omega X(\omega)^2\,d\mathbb P(\omega) -2m\int_\Omega X(\omega)\,d\mathbb P(\omega) +\int_\Omega m^2\,d\mathbb P(\omega). \end{align*} [guided] The point of this step is to replace the definition of variance, which involves the centered random variable $X-m$, by ordinary moments of $X$. We have already proved that the integrals involved are finite, so we may use linearity of the Lebesgue integral without encountering indeterminate expressions. For every $\omega\in\Omega$, expanding the square in $\mathbb R$ gives \begin{align*} (X(\omega)-m)^2=X(\omega)^2-2mX(\omega)+m^2. \end{align*} We must check that each term can be integrated separately. The function $\omega\mapsto X(\omega)^2$ is integrable by the finite second moment hypothesis. The function $\omega\mapsto X(\omega)$ is integrable by the previous step. The constant function $\omega\mapsto m^2$ is integrable because \begin{align*} \int_\Omega m^2\,d\mathbb P(\omega)=m^2\mathbb P(\Omega)=m^2<\infty. \end{align*} Therefore linearity of the Lebesgue integral applies to the pointwise identity above, giving \begin{align*} \operatorname{Var}(X) &=\int_\Omega (X(\omega)-m)^2\,d\mathbb P(\omega)\\ &=\int_\Omega X(\omega)^2\,d\mathbb P(\omega) -2m\int_\Omega X(\omega)\,d\mathbb P(\omega) +\int_\Omega m^2\,d\mathbb P(\omega). \end{align*} [/guided] [/step] [step:Use the probability normalization to obtain the computational formula] By the definition of $m$ and the fact that $\mathbb P(\Omega)=1$, \begin{align*} \int_\Omega X(\omega)\,d\mathbb P(\omega)=m, \qquad \int_\Omega m^2\,d\mathbb P(\omega)=m^2\mathbb P(\Omega)=m^2. \end{align*} Substituting these identities into the expression from the previous step gives \begin{align*} \operatorname{Var}(X) &=\mathbb E[X^2]-2m^2+m^2\\ &=\mathbb E[X^2]-m^2\\ &=\mathbb E[X^2]-(\mathbb E[X])^2. \end{align*} This proves the claimed computational formula for the variance. [/step]