Let f : [a,b] → R be a Riemann integrable function. Let α > 0 and β ∈ R. Then define g(x) := f(αx+β) on the interval I = [1/α(a−β), 1/α(b−β)]. Show that g is Riemann integrable on I
To show that g(x) is Riemann integrable on the interval I = [1/α(a-β), 1/α(b-β)], we need to prove that for any given ε > 0, there exists a partition P of I such that the upper Riemann sum and the lower Riemann sum of g(x) differ by at most ε.
Let's consider a partition P = {x_0, x_1, ..., x_n} of I, where x_0 = 1/α(a-β) and x_n = 1/α(b-β), such that each subinterval [x_{i-1}, x_i] has the same length.
Now, let's calculate the upper and lower Riemann sums for g(x) using this partition P.
The upper Riemann sum U(f, P) is given by:
U(f, P) = Σ(M_i Δx_i), for i = 1 to n,
where M_i is the supremum of g(x) in the subinterval [x_{i-1}, x_i], and Δx_i is the length of the subinterval [x_{i-1}, x_i].
Similarly, the lower Riemann sum L(f, P) is given by:
L(f, P) = Σ(m_i Δx_i), for i = 1 to n,
where m_i is the infimum of g(x) in the subinterval [x_{i-1}, x_i].
Now, let's analyze how g(x) behaves on the subintervals [x_{i-1}, x_i].
Since g(x) = f(αx + β) and x ranges from 1/α(a-β) to 1/α(b-β), we can rewrite the subinterval [x_{i-1}, x_i] in terms of x as [α(x_{i-1} - β), α(x_i - β)].
Let's denote the values of α(x_{i-1} - β) and α(x_i - β) as a_{i-1} and a_i, respectively.
The length of the subinterval [x_{i-1}, x_i] is given by:
Δx_i = a_i - a_{i-1}.
Now, g(x) = f(αx + β) becomes g(x) = f(a), where x ranges from a_{i-1} to a_i.
Since f(x) is Riemann integrable on the interval [a, b], it is also bounded. Therefore, there exist bounds M and m such that M ≥ f(x) ≥ m for all x in [a, b].
This means that for any x in the subinterval [a_{i-1}, a_i], we have m ≤ f(a) ≤ M, which implies that m ≤ g(x) ≤ M.
Now, let's calculate the difference between the upper and lower Riemann sums for g(x):
U(f, P) - L(f, P) = Σ(M_i Δx_i) - Σ(m_i Δx_i)
= Σ((M_i - m_i) Δx_i)
≤ M Σ(Δx_i) - m Σ(Δx_i)
= (M - m) Σ(Δx_i)
Since each subinterval [x_{i-1}, x_i] has the same length, Σ(Δx_i) is simply the total length of the interval I, which is 1/α(b-β) - 1/α(a-β) = 1/α(b-a).
Therefore, we have:
U(f, P) - L(f, P) ≤ (M - m) * (1/α(b-a))
Now, for any given ε > 0, we can choose a partition P such that (M - m) * (1/α(b-a)) < ε.
Hence, we have shown that for any ε > 0, there exists a partition P such that the difference between the upper and lower Riemann sums of g(x) on I is less than ε.
Therefore, g(x) is Riemann integrable on the interval I = [1/α(a-β), 1/α(b-β)].
To show that g is Riemann integrable on the interval I, we need to prove that the upper Riemann sum and the lower Riemann sum of g converge to the same value as the partition becomes finer.
First, let's define the partition. Since I is the interval [1/α(a−β), 1/α(b−β)], we can divide I into n subintervals of equal length by setting Δx = (1/α(b−β) - 1/α(a−β))/n. Then the partition points can be defined as x_i = 1/α(a−β) + iΔx for i = 0, 1, 2, ..., n.
Next, we consider the upper Riemann sum of g. It is defined as:
U(f, P) = Σ[(x_i - x_{i-1}) * sup(g(x))]
where sup(g(x)) is the supremum of g(x) on the ith subinterval.
In our case, g(x) = f(αx+β), so we have:
U(g, P) = Σ[(x_i - x_{i-1}) * sup(f(αx+β))]
Now, since f is Riemann integrable on [a, b], we know that it is bounded. Let M be an upper bound for f on [a, b]. Then we have:
sup(f(αx+β)) ≤ M
Therefore, we can rewrite the upper sum as:
U(g, P) ≤ Σ[(x_i - x_{i-1}) * M]
Now, let's consider the lower Riemann sum of g. It is defined as:
L(g, P) = Σ[(x_i - x_{i-1}) * inf(g(x))]
where inf(g(x)) is the infimum of g(x) on the ith subinterval.
Similarly, we have:
inf(f(αx+β)) ≥ m
where m is a lower bound for f on [a, b]. Therefore, the lower sum can be written as:
L(g, P) ≥ Σ[(x_i - x_{i-1}) * m]
Now, let's calculate the difference between the upper and lower Riemann sums:
U(g, P) - L(g, P) ≤ Σ[(x_i - x_{i-1}) * (M - m)]
Notice that (M - m) is a constant, so we can factor it out of the sum:
U(g, P) - L(g, P) ≤ (M - m) * Σ[(x_i - x_{i-1})]
But, Σ[(x_i - x_{i-1})] is just the total length of the interval I, which is (1/α(b−β) - 1/α(a−β)). Therefore, we have:
U(g, P) - L(g, P) ≤ (M - m) * (1/α(b−β) - 1/α(a−β))
As the partition becomes finer (i.e., as n tends to infinity), the length of each subinterval Δx tends to zero:
lim(n→∞) (1/α(b−β) - 1/α(a−β))/n = 0
Therefore, the difference between the upper and lower Riemann sums also tends to zero:
lim(n→∞) (U(g, P) - L(g, P)) ≤ (M - m) * 0 = 0
This shows that the upper and lower Riemann sums of g converge to the same value as the partition becomes finer. Hence, g is Riemann integrable on the interval I = [1/α(a−β), 1/α(b−β)].