Question

Suppose that \(f\) is integrable and \(g\) is bounded on \([a, b]\), and \(g\) differs from \(f\) only at points in a set \(H\) with the following property: For each \(\epsilon>0, H\) can be covered by a finite number of closed subintervals of \([a, b],\) the sum of whose lengths is less than \(\epsilon\). Show that \(g\) is integrable on \([a, b]\) and that $$ \int_{a}^{b} g(x) d x=\int_{a}^{b} f(x) d x $$

Short answer

Question: Prove that if \(f\) is integrable on \([a, b]\) and \(g\) is bounded on \([a, b]\) and differs from \(f\) only at points in a set \(H\) that can be covered by a finite number of closed subintervals with the sum of their lengths less than \(\epsilon\), then \(g\) is integrable on \([a, b]\) and \(\int_{a}^{b} g(x) dx = \int_{a}^{b} f(x) dx\). Answer: The proof consists of four steps: (1) using the definition of Riemann integrability for \(f\), (2) modifying the partition to cover the set \(H\), (3) exploring the connection between the Riemann sums for \(f\) and \(g\), and (4) showing the limits of the Riemann sums are equal. By following these steps, we can conclude that \(g\) is integrable on \([a, b]\) and \(\int_{a}^{b} g(x) dx = \int_{a}^{b} f(x) dx\).

Step-by-step solution

Use the definition of Riemann integrability for \(f\)

Given that \(f\) is integrable on \([a, b]\), for any \(\epsilon > 0\), there exists a partition \(P = \{a = x_0 < x_1 < \cdots < x_n = b\}\) of \([a, b]\) such that for every choice of points \(t_i \in [x_{i-1}, x_i]\), we have: $$ \left|\sum_{i=1}^{n} f(t_i)(x_i - x_{i-1}) - \int_a^b f(x) dx\right| < \frac{\epsilon}{2} $$

Modify the partition for \(H\)

Since \(H\) can be covered by a finite number of closed subintervals with the sum of their lengths less than \(\epsilon\), we can modify the partition \(P\) to include the endpoints of these subintervals. Let's call the new partition \(P'\). Since \(P'\) is a refinement of \(P\) (i.e., it contains all the points of \(P\) and possibly some new points), the same inequality holds for the Riemann sum for \(f\) with the partition \(P'\), i.e., $$ \left|\sum_{i=1}^{n'} f(t_i')(x_i' - x_{i-1}') - \int_a^b f(x) dx\right| < \frac{\epsilon}{2} $$ where \(P' = \{a = x_0' < x_1' < \cdots < x_{n'}' = b\}\), and \(t_i'\) lies in the interval \([x_{i-1}', x_i']\).

Connection between the sums for \(f\) and \(g\)

Now, let's consider the Riemann sum for \(g\) with the partition \(P'\). Since \(g\) differs from \(f\) only at points in the set \(H\), and \(H\) is covered within the subintervals added to the partition \(P\), we can bound the difference between the Riemann sums for \(f\) and \(g\) as follows: $$ \left|\sum_{i=1}^{n'} g(t_i')(x_i' - x_{i-1}') - \sum_{i=1}^{n'} f(t_i')(x_i' - x_{i-1}')\right| \leq \sum_{i=1}^{n'} |g(t_i') - f(t_i')|(x_i' - x_{i-1}') $$ Since \(g\) is bounded on \([a, b]\), there exists a constant \(M > 0\) such that \(|g(x) - f(x)| \leq M\) for all \(x \in [a, b]\). Therefore, we can bound the difference in the sums further: $$ \sum_{i=1}^{n'} |g(t_i') - f(t_i')|(x_i' - x_{i-1}') \leq M \sum_{i=1}^{n'} (x_i' - x_{i-1}') < M \epsilon $$ The last inequality follows from the property of the set \(H\).

Show the limits of the Riemann sums are equal

Now by adding the inequalities from Steps 1 and 3, we get: $$ \left|\sum_{i=1}^{n'} g(t_i')(x_i' - x_{i-1}') - \int_a^b f(x) dx\right| \leq \left|\sum_{i=1}^{n'} g(t_i')(x_i' - x_{i-1}') - \sum_{i=1}^{n'} f(t_i')(x_i' - x_{i-1}')\right| + \left|\sum_{i=1}^{n'} f(t_i')(x_i' - x_{i-1}') - \int_a^b f(x) dx\right| < M\epsilon + \frac{\epsilon}{2} $$ Since this inequality holds for any arbitrary \(\epsilon > 0\), it implies that \(g\) is integrable on \([a, b]\) and $$ \int_{a}^{b} g(x) dx = \int_{a}^{b} f(x) dx $$

Key concepts

Integrable Functions

Integrable functions are an important concept in calculus, particularly in understanding how we can compute the area under a curve. To say that a function is Riemann integrable over an interval \([a, b]\) means that we can calculate the integral, or the area under the curve, precisely using Riemann sums. The main condition for a function to be integrable is that the sum of the differences between the upper and lower sums can be made arbitrarily small by choosing a fine enough partition.

For a function \("f"\) to be integrable on an interval \([a, b]\), we first choose a partition \("P"\) of that interval. This means slicing it into subintervals. For each subinterval, we pick a point and form the Riemann sum, which approximates the integral of \("f"\). The function is Riemann integrable if this sum approaches the actual integral as the partition gets finer (i.e., as the intervals get smaller).

• \(f\) is integrable if, for every \(\epsilon > 0\), there is a partition, making the Riemann sum difference less than \(\epsilon\).
• \(g\) is said to be integrable under similar conditions as function \("f"\), allowing it to work with Riemann sums.

Bounded Functions

A function is bounded on a given interval if there exists a limit to the values the function can take over that interval. In simple terms, the function is not allowed to "run off" to infinity. For Riemann integrability, boundedness is crucial because unbounded functions on a closed interval \([a, b]\) can create difficulties in approximating the area under the curve.

Why does this matter? Well, the boundedness allows us to keep control over the function's behavior so that, even when calculating integrals, the values stay within a predictable range. This makes Riemann sums viable for calculating exact integrals, as seen in the exercise where \("g"\) stays bounded over \([a, b]\):

• If a function is bounded, \(\exists\) some constant \(M\) such that \(|g(x)| \leq M\) for all points \(x\) in the interval.
• Boundedness supports a consistent approximation of integrables, using defined limits.

Riemann Sum

The Riemann sum is a technique for estimating the total integral of a function. Imagine dividing the area under a curve into many "slabs" or rectangles. We then sum up the areas of these slabs to approximate the integral.

In a typical Riemann sum, you break the interval \([a, b]\) into \(n\) smaller intervals. For each of these intervals, you choose a sample point and calculate the area of each rectangle by multiplying the function value at the sample point by the width of the interval. The sum of all these small areas gives the approximate integral. As you increase \(n\) and narrow down each interval, the Riemann sums give values closer to the actual integral. This structure is useful for both \("f"\) and \("g"\) in your exercise:

• The finer the partition, the more accurate the Riemann sum becomes.
• Choosing the right sample point on each interval can help in reducing error further.
• The calculation closely depends on approximating rectangles' heights using the function's values at strategic points.