The question is this: Let f be an integrable function on [-a,a] (a>0). Assume f is an even function (i.e. f(-x) = f(x)) for all x ∈ [-a,a]. Then use the definition of an integral to prove $\int_{-a}^a f = 2\int_{0}^a f$. We need to show that U(f,P) = 2 U(f,P+) and L(f,P) = 2 L(f,P+). Could someone help with my proof?
My take on the proof: Consider a partition P+ = {0 = t0 < t1 < ... < tn = a} of [0, a] and a partition P_ = {-a = -tn < ... < -t1 < -t0 = 0} of [-a,0] such that U(f,P+) - L(f,P+) = ε/2 and U(f,P_) - L(f,P_) = ε/2.
Let P = P+ + P_ then P is a partition of [-a,a]. We have:
L(f) ≥ L(f, P) = L(f, P+) + L(f,P_) > U(f, P+) + U(f, P_) − ε and U(f) ≤ U(f, P) = U(f, P+) + U(f, P_) < L(f, P+) + L(f, P_) + ε
Because P+ is the inverse of P_, P = 2P+. Hence, L(f) ≥ L(f, P) = 2 L(f, P+) and U(f) ≥ U(f, P) = 2 U(f, P+). From here, I know that we need to show the inequality of 2 L(f,P+) ≤ $\int_{-a}^a f$ ≤ 2 U(f,P+) but I am not sure how to set it up. Could someone help me to check if what I did so far is correct and show how I can move on to proving this?