Question

(a) Prove: If \(\int_{a}^{b} f(x) d x\) exists, then for every \(\epsilon>0,\) there is a \(\delta>0\) such that \(\left|\sigma_{1}-\sigma_{2}\right|<\epsilon\) if \(\sigma_{1}\) and \(\sigma_{2}\) are Riemann sums of \(f\) over partitions \(P_{1}\) and \(P_{2}\) of \([a, b]\) with norms less than \(\delta\). (b) Suppose that there is an \(M>0\) such that, for every \(\delta>0,\) there are Riemann sums \(\sigma_{1}\) and \(\sigma_{2}\) over a partition \(P\) of \([a, b]\) with \(\|P\|<\delta\) such that \(\left|\sigma_{1}-\sigma_{2}\right| \geq\) \(M\). Use (a) to prove that \(f\) is not integrable over \([a, b]\).

Short answer

In summary, part (a) of the problem shows that if a function is integrable on an interval, then there exists a value of δ ensuring that the absolute difference between any two Riemann sums over partitions with norms less than δ is less than any given positive value ε. In part (b), we used the result from part (a) to prove that a function that does not satisfy this condition is not integrable over the given interval.

Step-by-step solution

Part (a): Proof for the existence of \(\delta\)

Given: \(\int_{a}^{b} f(x) d x\) exists. We are supposed to prove that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(\sigma_1\) and \(\sigma_2\) are Riemann sums of \(f\) over partitions \(P_1\) and \(P_2\) with norms less than \(\delta\), then the absolute difference between the two Riemann sums is less than \(\epsilon\). Now, we know that the definite integral of \(f(x)\) exists on \([a, b]\). This means that the limit of the Riemann sums (over partitions with mesh sizes approaching 0) converges to a unique value. Let this value be 'c'. Therefore, given any \(\epsilon > 0\), there exists a partition, say \(P^*\), with a mesh size less than \(\delta\) such that: \(| \sigma(P^*) - c | < \frac{\epsilon}{2}\), where \(\sigma(P^*)\) is the Riemann sum of \(f\) over partition \(P^*\). Now we know that the Riemann sums \(\sigma_1\) and \(\sigma_2\) over partitions \(P_1\) and \(P_2\) exist with norms less than \(\delta\). Then, we have: \(| \sigma_1 - c | < \frac{\epsilon}{2}\) and \(| \sigma_2 - c | < \frac{\epsilon}{2}\). Using the Triangle Inequality, we have: \(|\sigma_1 - \sigma_2| = |(\sigma_1 - c) - (\sigma_2 - c)| \leq |\sigma_1 - c| + |\sigma_2 - c| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\). Thus, for any \(\epsilon > 0\), there exists a \(\delta > 0\) such that the absolute difference between the two Riemann sums is less than \(\epsilon\).

Part (b): Proof for non-integrability of \(f\)

Given: For every \(\delta > 0\), there exist Riemann sums, \(\sigma_1\) and \(\sigma_2\), over a partition \(P\) of \([a, b]\) with \(\|P\| < \delta\) such that \(\left|\sigma_{1}-\sigma_{2}\right| \geq M\) for some \(M > 0\). We are going to use the result in part (a) to prove that function \(f\) is not integrable over \([a, b]\) under this given condition. Assume that \(f\) is integrable on \([a, b]\). Then, as per part (a), for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that the absolute difference between any two Riemann sums of \(f\) over partitions with norms less than \(\delta\) is less than \(\epsilon\). However, this result contradicts the given condition that for every \(\delta > 0\), there exist Riemann sums, \(\sigma_1\) and \(\sigma_2\), with norms less than \(\delta\) such that \(\left|\sigma_{1}-\sigma_{2}\right| \geq M\). As the conclusions reached from the integrability of \(f\) contradict the given condition, our hypothesis that \(f\) is integrable on \([a, b]\) must be false. Therefore, \(f\) is not integrable over \([a, b]\).

Key concepts

Definite Integral

Understanding the definite integral is foundational to grasping the broader concepts of calculus. A definite integral, represented as \(\int_{a}^{b} f(x) \, dx\), quantifies the net area under the curve of the function \(f(x)\) from point \(a\) to \(b\) on the x-axis. Picture it like summing up tiny slices of area under the curve, where each slice represents a function's value at a particular point, multiplied by an infinitesimally small width.

Moreover, the fundamental theorem of calculus ties the definite integral to the antiderivative, establishing that evaluating the antiderivative at the upper and lower bounds gives the same value as the definite integral—essentially linking the concept of accumulation (area under the curve) with the concept of rates of change (derivatives).

Partition of an Interval

Think of a partition of an interval like chopping a chocolate bar into pieces. Just as you break the chocolate into smaller segments, a partition of an interval \[a, b\] in mathematics involves dividing this interval into smaller subintervals. This is typically accomplished by choosing points \(a = x_0 < x_1 < x_2 < ... < x_{n-1} < x_n = b\).

Each chosen point is like a mark on the ruler that measures your chocolate bar—the interval from \(a\) to \(b\). Partitions are a crucial element in defining Riemann sums and are used to approximate the area under a curve. Different partitions can give rise to different Riemann sums, but as the norm of the partition gets smaller, these sums converge to the definite integral if the function is integrable.

Norm of a Partition

The norm of a partition gets into the nitty-gritty of how finely we chop our interval. It is defined as the width of the largest subinterval in a given partition of \[a, b\].

We denote the norm of a partition \(P\) by \(\|P\|\) and calculate it as \(\|P\| = \max\{x_{i+1} - x_i | i = 0, 1, 2, ... , n-1\}\). The goal is to make this norm as small as possible; this means more slicing, leading to a finer approximation of the area under the curve. When working with Riemann sums, we want the norm to be small (a small \(\delta\)) to ensure our sum accurately approximates the definite integral—the smaller the norm, the closer we get to the true area.

Triangle Inequality

The Triangle Inequality is a rule that's kind of like saying you can't shortcut through a park when walking the edges of its triangular fence—you have to walk two sides to get to the point directly across from you. In mathematical language, it states that for any real numbers \(a\) and \(b\), the absolute value of their sum is less than or equal to the sum of their absolute values: \( |a + b| \leq |a| + |b| \).

This property is massively important when proving concepts in analysis such as convergence and integrability. As in the exercise solution, we use the Triangle Inequality to show that the difference between two Riemann sums of a function on a given interval can be made arbitrarily small, an essential feature of functions that are integrable on that interval.