Define g : [0, 1] → R byg(x) =sin(x2)x if 0 < x ≤ 1,0 if x = 0.13. Prove that g ∈ R[0, 1]

No answer

To prove that g ∈ R[0, 1], we need to show that g is Riemann integrable on the interval [0, 1].

A function f is Riemann integrable on [a, b] if for every ε > 0, there exists a partition P of [a, b] such that U(P, f) - L(P, f) < ε, where U(P, f) and L(P, f) are the upper and lower Riemann sums of f with respect to P.

Step 1: Define the function g

We have g(x) = sin(x^2)/x for 0 < x ≤ 1 and g(x) = 0 for x = 0.

Step 2: Show that g is bounded

Since sin(x) is always between -1 and 1 for all x, and x^2/x = x is between 0 and 1 for 0 < x ≤ 1, we have -1 ≤ g(x) ≤ 1 for all x in [0, 1]. So g is bounded.

Step 3: Show that g is continuous almost everywhere

The function g is continuous on (0, 1] because the sine function and the function x^2/x = x are both continuous on this interval, and the quotient of two continuous functions is continuous wherever the denominator is not zero.

At x = 0, we have g(x) = 0 by definition. We can show that g is continuous at x = 0 by taking the limit as x approaches 0. Using L'Hopital's rule, we have

lim (x->0) sin(x^2)/x = lim (x->0) (2x cos(x^2))/1 = 0,

which is the value of g at x = 0. So g is continuous at x = 0.

Therefore, g is continuous everywhere on [0, 1] except possibly at the points where x^2 = nπ for some integer n, because the cosine function may be zero at these points. But these points form a countable set, and a function that is continuous except at a countable number of points is Riemann integrable.

Step 4: Show that g is Riemann integrable

Since g is bounded and continuous almost everywhere on [0, 1], it is Riemann integrable on this interval by Lebesgue's criterion for Riemann integrability.

Therefore, we have shown that g ∈ R[0, 1].

To prove that g ∈ R[0, 1], we need to show that g is Riemann integrable on the interval [0, 1].

A function f is Riemann integrable on [a, b] if for every ε > 0, there exists a partition P of [a, b] such that U(P, f) - L(P, f) < ε, where U(P, f) and L(P, f) are the upper and lower Riemann sums of f with respect to P.

Step 1: Show that g is bounded on [0, 1] The function g(x) = sin(x^2)/x is bounded on the interval (0, 1] because sin(x^2) is bounded between -1 and 1 for all x, and 1/x is also bounded on the interval (0, 1]. At x = 0, g(x) = 0, so g is also bounded at x = 0. Therefore, g is bounded on [0, 1].

Step 2: Show that g is continuous almost everywhere on [0, 1] The function g(x) = sin(x^2)/x is continuous for all x in (0, 1] because both sin(x^2) and 1/x are continuous on this interval, and the quotient of two continuous functions is continuous wherever the denominator is not zero. At x = 0, g(x) = 0, so g is discontinuous at x = 0. However, the set of discontinuities of a bounded function on [a, b] is a null set if it consists of a finite number of points, so the set of discontinuities of g on [0, 1] is a null set.

Step 3: Apply the Lebesgue's integrability criterion Since g is bounded on [0, 1] and the set of its discontinuities on [0, 1] is a null set, by Lebesgue's integrability criterion, g is Riemann integrable on [0, 1]. Therefore, g ∈ R[0, 1].