[proofplan] We first verify that the affine transform $aX+b$ has finite second moment, so its variance is defined. Then we compute its expectation using linearity of the [Lebesgue integral](/page/Lebesgue%20Integral) with respect to the probability measure $\mathbb P$. Finally, we center $aX+b$ and observe that the constant shift $b$ cancels, leaving a factor $a$ inside the square and hence a factor $a^2$ outside the expectation. [/proofplan] [step:Verify that the affine transform is square-integrable] Define the [random variable](/page/Random%20Variable) \begin{align*} Y:(\Omega,\mathcal F)&\to(\mathbb R,\mathcal B(\mathbb R))\\ \omega&\mapsto aX(\omega)+b. \end{align*} Since $X$ is measurable and affine maps $\mathbb R\to\mathbb R$ are Borel-measurable, $Y$ is a real-valued random variable. In this proof, for any integrable real-valued random variable $Z: (\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$, the expectation is \begin{align*} \mathbb E[Z]:=\int_\Omega Z(\omega)\,d\mathbb P(\omega), \end{align*} and for any square-integrable real-valued random variable $Z$, its variance is \begin{align*} \operatorname{Var}(Z):=\mathbb E\left[(Z-\mathbb E[Z])^2\right]. \end{align*} For every $\omega\in\Omega$, applying the elementary inequality $|u+v|^2\le 2|u|^2+2|v|^2$ with $u=aX(\omega)$ and $v=b$ gives \begin{align*} |Y(\omega)|^2 =|aX(\omega)+b|^2 \le 2a^2|X(\omega)|^2+2b^2. \end{align*} Integrating this inequality with respect to $\mathbb P$ gives \begin{align*} \mathbb E[Y^2] =\int_\Omega |Y(\omega)|^2\,d\mathbb P(\omega) \le 2a^2\int_\Omega |X(\omega)|^2\,d\mathbb P(\omega)+2b^2\mathbb P(\Omega) <\infty, \end{align*} because $\mathbb E[X^2]<\infty$ and $\mathbb P(\Omega)=1$. Hence $Y=aX+b$ is square-integrable. [/step] [step:Compute the expectation of $aX+b$] Since $X$ is square-integrable, it is integrable: for every $\omega\in\Omega$, \begin{align*} |X(\omega)|\le \frac{1+|X(\omega)|^2}{2}, \end{align*} and the right-hand side has finite $\mathbb P$-integral. Therefore $\mathbb E[X]$ is finite. By the linearity theorem for the Lebesgue integral applied with respect to the measure $\mathbb P$, \begin{align*} \mathbb E[Y] &=\int_\Omega \bigl(aX(\omega)+b\bigr)\,d\mathbb P(\omega)\\ &=a\int_\Omega X(\omega)\,d\mathbb P(\omega)+b\int_\Omega 1\,d\mathbb P(\omega)\\ &=a\mathbb E[X]+b. \end{align*} [guided] We need the expectation of $Y=aX+b$ before we can compute the variance, because variance is defined by centering around the expectation. First, $\mathbb E[X]$ is finite. Indeed, for every $\omega\in\Omega$, \begin{align*} |X(\omega)|\le \frac{1+|X(\omega)|^2}{2}. \end{align*} The function on the right has finite integral with respect to $\mathbb P$, since $\mathbb P(\Omega)=1$ and $\mathbb E[X^2]<\infty$. Hence $X$ is integrable. Now apply the linearity theorem for the Lebesgue integral with respect to $\mathbb P$ to the integrable function $X:\Omega\to\mathbb R$ and the constant function $\omega\mapsto b$. We obtain \begin{align*} \mathbb E[Y] &=\int_\Omega Y(\omega)\,d\mathbb P(\omega)\\ &=\int_\Omega \bigl(aX(\omega)+b\bigr)\,d\mathbb P(\omega)\\ &=a\int_\Omega X(\omega)\,d\mathbb P(\omega)+b\int_\Omega 1\,d\mathbb P(\omega)\\ &=a\mathbb E[X]+b\mathbb P(\Omega)\\ &=a\mathbb E[X]+b. \end{align*} The important point is that the additive constant contributes exactly $b$ to the expectation because $\mathbb P$ is a probability measure. [/guided] [/step] [step:Center the affine transform and factor out the scalar] Using the expectation computed above, for every $\omega\in\Omega$ we have \begin{align*} Y(\omega)-\mathbb E[Y] &=aX(\omega)+b-\bigl(a\mathbb E[X]+b\bigr)\\ &=a\bigl(X(\omega)-\mathbb E[X]\bigr). \end{align*} Therefore, by the definition of variance stated above, \begin{align*} \operatorname{Var}(Y) &=\mathbb E\left[\bigl(Y-\mathbb E[Y]\bigr)^2\right]\\ &=\int_\Omega \bigl(Y(\omega)-\mathbb E[Y]\bigr)^2\,d\mathbb P(\omega)\\ &=\int_\Omega a^2\bigl(X(\omega)-\mathbb E[X]\bigr)^2\,d\mathbb P(\omega)\\ &=a^2\int_\Omega \bigl(X(\omega)-\mathbb E[X]\bigr)^2\,d\mathbb P(\omega)\\ &=a^2\operatorname{Var}(X), \end{align*} where the penultimate equality uses the linearity theorem for the Lebesgue integral for the constant scalar $a^2$, and the last equality uses the definition of $\operatorname{Var}(X)$. Since $Y=aX+b$, this proves \begin{align*} \operatorname{Var}(aX+b)=a^2\operatorname{Var}(X). \end{align*} [/step]