Show that π(βπ, π) = βπΏ(π, π) and πΏ(βπ, π) = βπ(π, π)
Suppose that f is integrable on [a; b]. For "\\in > 0" , there exist partitions P1 and P2 such that
"L(f, P_1) > L(f) \u2212 \\frac{\\in}{2} \\space \nand \\space U(f; P2) < U(f) + \\frac{\\in}{2}\\\\\nFor \\space P := P_1 \u222a P_2 \\space we \\space have\\\\\nL(f) \u2212\\frac{\\in}{2} \n< L(f, P_1) \u2264 L(f; P ) \u2264 U(f, P ) \u2264 U(f, P_2) < U(f) + \\frac{\\in}{2} \\\\"
Since L(f) = U(f), it follows that "U(f, P ) - L(f,P )< \\in"
Implies π(βπ, π) = βπΏ(π, π) and πΏ(βπ, π) = βπ(π, π)
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