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  "body": "\n## Proof\n\nWe prove the formula for nonnegative parts by approximation with simple functions and monotone convergence, then combine the parts.\n\n1. Let $X$ and $Y$ be [[independent|0E9502NT92]] integrable real [[random variables|0DKRR22AEQ]] on a probability space $(\\Omega, \\mathcal{F}, P)$.\n2. For a real number $u$ put $u^+ = \\max(u, 0)$ and $u^- = \\max(-u, 0)$. Then $u = u^+ - u^-$ and $\\abs{u} = u^+ + u^-$.\n3. For $n \\ge 1$ and $u \\ge 0$ let $\\varphi_n(u) = \\min\\left(n, 2^{-n} \\floor{2^n u}\\right)$, where $\\floor{\\cdot}$ is the [[floor function|0YVRNEDSVG]].\n4. The function $\\varphi_n$ takes only the finitely many values $k 2^{-n}$ with $0 \\le k \\le n 2^n$. It is nondecreasing, and $0 \\le \\varphi_n(u) \\le u$.\n5. We have $\\varphi_n(u) \\le \\varphi_{n+1}(u)$. Indeed $2 \\floor{2^n u}$ is an integer at most $2^{n+1} u$, so it is at most $\\floor{2^{n+1} u}$.\n6. Dividing by $2^{n+1}$ gives $2^{-n} \\floor{2^n u} \\le 2^{-n-1} \\floor{2^{n+1} u}$. Since also $n \\le n + 1$, the minimum in step 3 does not decrease.\n7. We have $\\varphi_n(u) \\to u$. For $n \\ge u$ we get $\\varphi_n(u) = 2^{-n} \\floor{2^n u}$, and this lies in $(u - 2^{-n}, u]$.\n8. Let $g$ be one of the maps $x \\mapsto x^+$ and $x \\mapsto x^-$ from $\\R$ to $[0, \\infty)$. Each is monotone, so $\\varphi_n \\circ g$ is monotone on $\\R$.\n9. Hence for each $c$ the set $\\set{x \\in \\R : \\varphi_n(g(x)) = c}$ is an [[interval|0TK8Y3J5RW]] or empty. If $x \u003c z \u003c y$ and the values at $x$ and $y$ equal $c$, monotonicity forces the value $c$ at $z$.\n10. Every interval is a [[Borel set|0BVHZXWKCD]]. It is open or closed, or it is an open interval together with one closed endpoint.\n11. Let $g$ and $h$ be maps as in step 8, let $U = g(X)$ and $V = h(Y)$, and assume $E[U] \u003c \\infty$ and $E[V] \u003c \\infty$. We show $E[UV] = E[U] \\, E[V]$ in steps 12 to 17.\n12. Let $U_n = \\varphi_n(U)$, with distinct values $c_1, \\dots, c_r$. By steps 9 and 10, $A_i = \\set{U_n = c_i} = \\set{X \\in J_i}$ with Borel sets $J_i$. The $A_i$ are disjoint events covering $\\Omega$.\n13. In the same way, $V_n = \\varphi_n(V)$ has distinct values $d_1, \\dots, d_s$, taken on the disjoint events $B_j = \\set{Y \\in K_j}$ with Borel sets $K_j$.\n14. So $U_n V_n = \\sum_{i, j} c_i d_j \\mathbf{1}_{A_i \\cap B_j}$ is a nonnegative [[simple function|7WMBX7M4P6]], written with pairwise disjoint sets $A_i \\cap B_j$ and [[indicator functions|0VSB0XF4G7]].\n15. By the [[integral of a non-negative function|GEA8H3228S]] for simple functions, $E[U_n V_n] = \\sum_{i, j} c_i d_j \\, P(A_i \\cap B_j)$. By independence, $P(A_i \\cap B_j) = P(X \\in J_i) \\, P(Y \\in K_j) = P(A_i) \\, P(B_j)$.\n16. So $E[U_n V_n] = \\left(\\sum_i c_i P(A_i)\\right) \\left(\\sum_j d_j P(B_j)\\right) = E[U_n] \\, E[V_n]$, again by the formula for simple functions.\n17. By steps 5 and 7, $U_n$ increases to $U$ and $V_n$ increases to $V$. Products of nondecreasing nonnegative sequences are nondecreasing, so $U_n V_n$ increases to $UV$.\n18. By the [[monotone convergence theorem|G2S6H789RG]], $UV$ is measurable, and $E[U_n] \\to E[U]$, $E[V_n] \\to E[V]$ and $E[U_n V_n] \\to E[UV]$.\n19. The limits $E[U]$ and $E[V]$ are finite. By the [[limit of a product of sequences|0734RSVF2V]] and step 16, $E[UV] = E[U] \\, E[V]$. This proves the claim of step 11.\n20. Now $X^+ \\le \\abs{X}$, so every simple $s$ with $0 \\le s \\le X^+$ also satisfies $s \\le \\abs{X}$. By the supremum definition, $E[X^+] \\le E[\\abs{X}] \u003c \\infty$.\n21. In the same way $E[X^-]$, $E[Y^+]$ and $E[Y^-]$ are finite, because $X$ and $Y$ are integrable.\n22. For signs $\\sigma, \\tau \\in \\set{+, -}$ let $W_{\\sigma\\tau} = X^\\sigma Y^\\tau$. By steps 11 to 21, each $W_{\\sigma\\tau}$ is a nonnegative random variable with $E[W_{\\sigma\\tau}] = E[X^\\sigma] \\, E[Y^\\tau] \u003c \\infty$. So each is integrable.\n23. Expanding $(X^+ - X^-)(Y^+ - Y^-)$ gives $XY = W_{++} - W_{+-} - W_{-+} + W_{--}$ at every point of $\\Omega$.\n24. By [[linearity of expectation|0P7PTC8DDE]], applied three times, $XY$ is integrable and $E[XY] = E[W_{++}] - E[W_{+-}] - E[W_{-+}] + E[W_{--}]$.\n25. By step 22, the right side is $E[X^+] E[Y^+] - E[X^+] E[Y^-] - E[X^-] E[Y^+] + E[X^-] E[Y^-]$.\n26. This factors as $\\left(E[X^+] - E[X^-]\\right)\\left(E[Y^+] - E[Y^-]\\right)$. By the definition of [[expected value|0DCT0QZVQ1]], it equals $E[X] \\, E[Y]$.\n\n## Idea\n\nFor simple random variables built from events of the form $\\set{X \\in J}$ and $\\set{Y \\in K}$, independence turns the expectation of the product into a double sum that factors. Every nonnegative random variable is an increasing limit of such simple ones, and monotone convergence carries the identity to the limit. Splitting $X$ and $Y$ into positive and negative parts handles the general integrable case.\n",
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