# Question: If f A R is non negative

If f: A→R is non-negative and ∫ Af = 0, show that B = {x: f (x) ≠ 0} has measure 0.

## Answer to relevant Questions

Let U be the open set of Problem 3-11. Show that if f = X except on a set of measure 0, then f is not integrable on [0, 1]Show by induction on n that R = [a1, b1] x..x [an, bn] is not a set of measure 0 (or content 0) if ai < bi for each i.If f: [a, b] x [c, d] → R is continuous and D2f is continuous, define F (x, y) = ∫xa (t,y) dt a. Find D1F and D2F. (b) If G (x) = ∫ g(x) f (t, x) dt, find G1 (x).Find a counter-example to Theorem 5-2 if condition (3) is omitted. Following the hint, consider f: (- 2π, 2π) →R2 defined byIf M is an -dimensional manifold-with-boundary in Rn, define μ as the usual orientation of M x = Rnx (the orientation μ so defined is the usual orientation of M. If xЄ∂M, show that the two definitions ...Post your question