Prove that the function defined by:
f(x)={1 if x is rational, 0 if x is irrational
is not integrable on [0,1]. Show that no matter how small the norm of the partition, ||P||, the Riemann sum can be made to have value either 0 or 1.
To prove that the function f(x) defined as:
f(x) = {1 if x is rational, 0 if x is irrational}
is not integrable on the interval [0, 1], we can use the fact that a function is integrable if and only if the upper and lower Riemann sums converge to the same value as the partition norm approaches zero.
Let's assume, for contradiction, that the function f(x) is integrable on [0, 1]. Then, for any partition P = {x0, x1, x2, ..., xn}, with ||P|| being the norm of the partition (maximum width of the subintervals), the upper Riemann sum U(f, P) and lower Riemann sum L(f, P) should converge to the same value as ||P|| approaches zero.
Consider any partition P = {x0, x1, x2, ..., xn} of [0, 1], and let Mj and mj be the supremum and infimum of f(x) on the subinterval [xj-1, xj] respectively.
Since the function f(x) takes the value 1 when x is rational and 0 when x is irrational, we have Mj = 1 for all j. Therefore, the upper Riemann sum U(f, P) is given by:
U(f, P) = Σ (Mj * Δxj)
= Σ (1 * Δxj)
= Σ Δxj
= 1
where Δxj = xj - xj-1 is the width of the subinterval.
Now, let's consider the lower Riemann sum L(f, P). Since the function f(x) takes the value 0 on the irrational numbers in each subinterval, the infimum of f(x) for each subinterval is 0. Hence, we have:
L(f, P) = Σ (mj * Δxj)
= Σ (0 * Δxj)
= Σ 0
= 0
Therefore, we have U(f, P) = 1 and L(f, P) = 0 for any partition P.
Now, as the norm of the partition ||P|| approaches zero, the width of each subinterval Δxj also approaches zero. However, no matter how small we choose the norm of the partition, we can always find both rational and irrational numbers within each subinterval. This implies that for any partition, there will always be subintervals containing both rational and irrational numbers, leading to a Riemann sum value of 1 and 0.
Hence, in this case, the upper and lower Riemann sums do not converge to the same value as the partition norm approaches zero. Therefore, by contradiction, the function f(x) defined as:
f(x) = {1 if x is rational, 0 if x is irrational}
is not integrable on the interval [0, 1].