Final answer:
To prove that L(f) ≤ L(g) and U(f) ≤ U(g), we need to show that the lower and upper sums of f are less than or equal to the lower and upper sums of g, respectively. This can be demonstrated by considering the infimum and supremum of f and g on each subinterval of a partition of I := [a, b].
Step-by-step explanation:
To prove that L(f) ≤ L(g) and U(f) ≤ U(g), we need to show that the lower and upper sums of f are less than or equal to the lower and upper sums of g, respectively. Since f(x) ≤ g(x) for all x ∈ I := [a, b], we can see that for each partition P of I, the lower sum and upper sum for f, denoted as L(f, P) and U(f, P), respectively, will be less than or equal to the lower sum and upper sum for g, denoted as L(g, P) and U(g, P), respectively.
For example, let's consider a specific partition P of I. Since f(x) ≤ g(x) for all x ∈ I, it follows that the infimum of f on each subinterval of P will be less than or equal to the infimum of g on that subinterval. Therefore, L(f, P) ≤ L(g, P) for any partition P of I, which means L(f) ≤ L(g). Similarly, we can show that U(f) ≤ U(g) using the supremum of f and g on each subinterval of P.