Question

(i) Prove that if \(f\) is constant \((f=c \neq \pm \infty)\) on \(I-Q,\) then $$ \int_{a}^{b} f=(b-a) c \quad \text { for } a, b \in I $$ (ii) Hence prove that if \(f=c_{k} \neq \pm \infty\) on $$ I_{k}=\left[a_{k}, a_{k+1}\right), \quad a=a_{0}

Short answer

Integrals for piecewise constant functions on intervals excluding rationals sum to total integral.

Step-by-step solution

Understand the Problem Statement

We need to prove two things: (i) If the function \(f\) is constant over an interval except for some points (rationals not affecting the value of the integral), then its integral over that interval equals the length of the interval times the constant. (ii) For a function that is piecewise constant on sub-intervals, show that its integral over the entire interval is the sum of the integrals over each sub-interval. Additionally, prove the result holds even if each sub-interval has some exceptional points.

Prove (i) for Constant Function over Irrationals

Let \(f(x) = c\) for all \(x \in I \setminus Q\). Since the irrational numbers \(I \setminus Q\) are dense in \(I\), for the Riemann integral, the finite choice of partition points, which are Rational, does not affect the integral. The contribution of the Rational points set measure \(0\) cannot change the integral:\[\int_a^b f(x) \, dx = \int_a^b c \, dx = c \cdot (b-a) = (b-a) c\].

Understand Interval Partition for (ii)

The function \(f\) is a piecewise constant, with each constant on sub-intervals \([a_k, a_{k+1})\). The aim is to express the total integral over \([a, b]\) as a sum of integrals over these sub-intervals.

Compute Integral Over Each Sub-Interval

For each sub-interval \([a_k, a_{k+1})\), \(f(x) = c_k\), so the integral is:\[\int_{a_k}^{a_{k+1}} f(x) \, dx = c_k \cdot (a_{k+1} - a_k)\].Since the Rationals do not affect the resulting integral, this representation is maintained.

Sum the Integrals Over All Sub-Intervals

Add the integrals computed for all sub-intervals to find the total integral over \([a, b]\): \[\int_a^b f(x) \, dx = \sum_{k=0}^{n-1} \int_{a_k}^{a_{k+1}} f(x) \, dx = \sum_{k=0}^{n-1} (a_{k+1} - a_k) c_k\].

Include Exceptional Sets and Show Equivalence

Even if \(f\) is constant over \(I_k \setminus Q_k\) instead of \(I_k\), due to the Rationals having measure zero, using Riemann integration, an exceptional set of Rationals does not change the value of the integrals, maintaining the result for \(f\) being constant over \(I_k \setminus Q_k\).

Key concepts

Piecewise constant function

A piecewise constant function is a type of function that can be broken down into a series of sub-intervals, where the function maintains a constant value within each sub-interval. This is a useful concept in mathematics, especially when calculating integrals, as it simplifies the process of integration by reducing it to handling a constant value over each part of the interval.

For example, consider a function defined on the interval \([a, b]\) which is divided into sub-intervals \([a_k, a_{k+1})\). On each of these sub-intervals, the function takes on a constant value \(c_k\). This means the function is not continuous but rather jumps between different constant levels as you move from one sub-interval to the next.

• Such functions can make complex problems easier to solve by turning them into a series of simple calculations involving constants.
• Applications of piecewise constant functions can often be found in signal processing and time series analysis where a signal is constant over time periods.

Dense sets

A dense set is an important concept in real analysis. A set is dense in a given interval if between any two numbers within that interval, you can find an element from the dense set. This suggests that the elements of the set are packed closely together, with no gaps.

For example, the rational numbers are dense in the real numbers. This means that between any two real numbers, no matter how close together, you can always find a rational number. For piecewise constant functions on intervals, this property of dense sets is significant because:

• It implies every real interval contains points of the dense set, such as the irrationals in an interval of real numbers.
• In the context of Riemann integration, it assures us that having dense sets of irrationals between partition points means the constant function's value holds over the entire interval.

This ensures that the finite number of partitions used in Riemann integration does not skip over any significant "gaps," solidifying the accuracy of the integral even if exceptions exist, like isolated rational points.

Measure zero

The concept of measure zero is fundamental in analysis and allows us to simplify the integration of functions with exceptional points, like those affecting rational numbers. A set having measure zero means its "size" in terms of length or area is zero, though it may have infinitely many elements.

An intuitive example is the set of rational numbers within any interval of real numbers. Despite there being infinitely many rationals within any segment of the real line, their influence on the length of intervals they reside in is none, literally zero measure.

• This property allows us to ignore these points when calculating integrals.
• For Riemann integration, this means that any partitioning or individual rational values within the interval do not contribute to the integration outcome, ensuring focus remains on the constant intervals where the function is well-defined.

Thus, when handling piecewise constant functions, sets of measure zero (like any isolated rational subsets) can often be ignored, simplifying the integration process.