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Problem 187

# Given that a bounded function $$\mathrm{f}(\mathrm{x})$$ is Riemann integrable in $$[a, b]$$ if and only if given any $$\varepsilon>0$$ there exists a partition with upper and lower sums $$\mathrm{U}$$ and $$\mathrm{L}$$ such that $$\mathrm{U}-\mathrm{L}<\varepsilon$$, prove that a continuous function $$\mathrm{f}(\mathrm{x})$$ in $$[\mathrm{a}, \mathrm{b}]$$ is Riemann integrable in $$[\mathrm{a}, \mathrm{b}]$$.

Expert verified
To prove that a continuous function $$f(x)$$ in $$[a, b]$$ is Riemann integrable, follow these steps: 1. Understand that a continuous function on $$[a, b]$$ is uniformly continuous, so for every $$\varepsilon>0$$, there is a $$\delta>0$$ for which $$|x-y|<\delta$$ implies $$|f(x)-f(y)|<\frac{\varepsilon}{b-a}$$. 2. Divide the interval $$[a, b]$$ into equally spaced subintervals using a partition with $$\Delta x = \frac{b-a}{n}<\delta$$. 3. Find the minimum and maximum values of $$f(x)$$ in each subinterval. 4. Calculate the lower sums $$L$$ and upper sums $$U$$ using these minimum and maximum values. 5. Show that $$U-L<\varepsilon$$ by observing that $$U-L \leq \Sigma \frac{\varepsilon}{b-a} * \Delta x =\varepsilon$$. 6. Conclude that the continuous function $$f(x)$$ is Riemann integrable, as we have shown that there exists a partition with upper and lower sums $$\mathrm{U}$$ and $$\mathrm{L}$$ such that $$U-L<\varepsilon$$.
See the step by step solution

## Step 1: Understand the definitions and properties of continuous functions and Riemann integrable functions.

A continuous function is a function whose graph has no breaks or gaps. A function is Riemann integrable if and only if for every ε>0, there exists a partition with upper and lower sums U and L such that U-L<ε. A continuous function on a closed interval [a, b] is uniformly continuous. Uniform continuity states that for every ε>0, there exists a δ>0 such that if the distance between any two points x and y in [a, b] is less than δ, then the distance between f(x) and f(y) is less than ε.

## Step 2: Divide the interval [a, b] into equally spaced subintervals.

Choose a partition of the interval [a, b] such that the length of each partition is: Δx = (b - a) / n, where n is a positive integer. Since the function is uniformly continuous, there must exist a δ>0 such that, whenever |x-y|<δ, |f(x)-f(y)|<ε/(b-a). Now, select n sufficiently large such that: Δx = (b - a) / n < δ.

## Step 3: Find upper and lower bounds for each subinterval.

For each subinterval [x_i, x_(i+1)], find the minimum value m_i and the maximum value M_i of the function f(x) in that subinterval. Since f(x) is continuous, the minimum and maximum values exist.

## Step 4: Calculate the lower sum L and the upper sum U.

Using the m_i and M_i found in step 3, calculate the lower sum L and the upper sum U for the partition as follows: L = Σ (m_i * Δx) for i=1 to n, U = Σ (M_i * Δx) for i=1 to n. Here, the symbol Σ denotes taking the sums of the products for each subinterval.

## Step 5: Show that U-L < ε.

Using the results of steps 2 and 4, observe that the difference U-L for the partition can be derived as follows: U - L = Σ (M_i * Δx) - Σ (m_i * Δx) = Σ (M_i - m_i) * Δx Since |f(x)-f(y)|<ε/(b-a) for all x and y in the same subinterval, we have that: U - L ≤ Σ ε/(b-a) * Δx = (ε/(b-a)) * Σ Δx = (ε/(b-a)) * (b - a) = ε This means that for the given partition and any ε>0, U-L < ε.

## Step 6: Conclude that the continuous function f(x) is Riemann integrable.

Since for any ε>0, there exists a partition with upper and lower sums U and L such that U-L<ε, it follows that the continuous function f(x) in [a, b] is Riemann integrable in [a, b].

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