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Problem 185

Let $$\mathrm{f}(\mathrm{x})=1, \mathrm{x}$$ rational and $$=0, \mathrm{x}$$ rational be defined in the interval $$[a, b]$$. Show that is not Riemann integrable

Expert verified
The function $$f(x)$$ is not Riemann integrable in the interval $$[a, b]$$, as it does not satisfy the condition that the limit of the difference between the upper and lower sums tends to 0 as the partition's mesh size tends to 0. Instead, we found that the difference between the upper and lower sums is $$b-a$$, which is a non-zero value.
See the step by step solution

Step 1: Understanding the function

Given function f(x) is defined as: $$f(x) = \begin{cases} 1, &\text{if }x\text{ is rational}, \\ 0, &\text{if }x\text{ is irrational} \end{cases}$$ The function is a "step" function, taking values 1 when x is rational and 0 when x is irrational. We will analyze the function in the interval [a, b].

Step 2: Partitioning the interval [a, b]

Consider a partition P of the interval [a, b] with n subintervals. Each subinterval contains a rational number and an irrational number since they are dense in every subinterval.

Step 3: Calculate the lower sum

Since the function has a value of 0 for irrational numbers, we can take the infimum of the function to be 0 in each subinterval. Therefore, the lower sum for the given function in every partition is always 0: $$L(P, f) = 0$$

Step 4: Calculate the upper sum

Since there are rational numbers in each subinterval (as rational numbers are dense), we can take the supremum of the function to be 1 in each subinterval. Therefore, the upper sum for each given partition is the width of the interval multiplied by the number of subintervals: $$U(P, f) = (b - a)$$

Step 5: Determine if the Riemann sum limit exists

The function f(x) would be Riemann integrable if, as the mesh size of the partitions tends to 0, the difference between the upper and lower sums tends to 0: $$\lim_{||P||\rightarrow 0} (U(P, f) - L(P, f)) = 0$$ However, using our calculations from steps 3 and 4, we have: $\lim_{||P||\rightarrow 0} ((b-a)-0) = b-a$ Since the difference between the upper and lower sums is non-zero, the function f(x) is not Riemann integrable in the interval [a, b].

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