Problem 3
Question
3-3. Let \(f, g: A \rightarrow \mathbf{R}\) be integrable. (a) For any partition \(P\) of \(A\) and subrectangle \(S\), show that $$ \begin{aligned} m_{S}(f)+m_{S}(g) \leq m_{S}(f+g) \quad \text { and } \quad M_{S}(f+g) \\ & \leq M_{s}(f)+M_{S}(g) \end{aligned} $$ and therefore $$ \begin{aligned} L(f, P)+L(g, P) \leq L(f+g, P) \quad \text { and } \quad & U(f+\theta, P) \\ & \leq U(f, P)+U(g, P) \end{aligned} $$ (b) Show that \(f+g\) is integrable and \(\int_{A} f+g=\int_{A} f+\int_{A} g\). (c) For any constant \(c\), show that \(\int_{A} c f=c \int_{A} f\).
Step-by-Step Solution
Verified Answer
The function sums and scalars preserve integrability, confirming \( \int_{A} f+g = \int_{A} f + \int_{A} g \) and \int_{A} c f = c \int_{A} f.
1Step 1 - Definitions and Background
Given two integrable functions, \( f \) and \( g \), we need to show certain properties about their lower sums \( L \) and upper sums \( U \). These sums relate to integrability and the Riemann sums for the integral over \( A \).
2Step 2 - Analyzing \( m_S \) and \( M_S \)
For any subrectangle \( S \) in a partition \( P \) of \( A \), the infimum (minimum bound) and supremum (maximum bound) over a subrectangle are given by \( m_S(f) \), \( M_S(f) \), respectively. Similarly, define \( m_S(g) \) and \( M_S(g) \) for \( g \).
3Step 3 - Proving Inequalities for \( m_S \) and \( M_S \)
By properties of infimum and supremum for sums, \[ m_S(f) + m_S(g) \leq m_S(f+g) \] and \[ M_S(f+g) \leq M_S(f) + M_S(g) \]. Here, \( f+g \) are the sum of the functions. Hence, \( m_S \) and \( M_S \) inequalities hold for any partition.
4Step 4 - Showing Integrability and Equality of Integrals
Because the inequalities hold for the lower sums \( L \) and upper sums \( U \), summing over all subrectangles in the partition validates: \[ L(f, P) + L(g, P) \leq L(f+g, P) \] and \[ U(f+g, P) \leq U(f, P) + U(g, P) \]. Thus, \( f+g \) maintains proper bounds indicating it is integrable with \[ \int_{A} f+g = \int_{A} f + \int_{A} g \].
5Step 5 - Scaling Functions by Constants
For any constant \( c \), let \( cf \) be the scaled function. Note that scaling a function multiplies its values, thus: \[ L(cf) = cL(f) \] and \[ U(cf) = cU(f) \]. This implies: \[ \int_{A} cf = c \int_{A} f \].
Key Concepts
Integrable FunctionsRiemann SumsInfimum and SupremumLower SumUpper Sum
Integrable Functions
An integrable function is one that satisfies the conditions needed to compute its integral over a given domain. In simple terms, if we can find a value that represents the area under the curve of a function, that function is considered integrable. For functions defined on a set A, like f and g in the exercise, integrability means they have well-defined integrals over A. These integrals exist if we can approximate the area under the curves using techniques like Riemann sums.
Riemann Sums
Riemann sums are a method for approximating the integral of a function. They involve dividing the domain of the function into smaller sub-intervals, calculating the function's value at specific points within these intervals, and then summing these values, each multiplied by the width of the sub-interval. This sum approximates the total area under the curve. The formula for a Riemann sum for a partition P is: \[ R(f, P) = \sum_{i=1}^{n} f(x_i^*) \Delta x_i \] Here, \( x_i^* \) is a point in the sub-interval, and \( \Delta x_i \) is the width of the sub-interval. As the width of the intervals approaches zero, the Riemann sum converges to the exact integral of the function. Definitions of lower and upper sums (LorRiemann sums) use infimum and supremum values within each sub-interval.
Infimum and Supremum
The infimum and supremum of a function over a set are key concepts in defining lower sums and upper sums. The infimum of a function over a set is the greatest value that is less than or equal to every function value on that set. It represents a lower bound. The supremum is the least value that is greater than or equal to every function value, acting as an upper bound. For a function f over a subrectangle S, these are defined as: \( m_S(f) = \text{inf} \{ f(x) : x \in S \} \) and \( M_S(f) = \text{sup} \{ f(x) : x \in S \} \). These bounds help us define and calculate lower and upper sums, thus affecting how we approximate integrals.
Lower Sum
The lower sum is an approximation of the integral of a function using the infimum of the function values in each sub-interval. For a partition P of a domain A, the lower sum L is calculated by: \[ L(f, P) = \sum_{S \in P} m_S(f) \cdot \text{Area}(S) \] Here, \( m_S(f) \) is the infimum of f in the subrectangle S, and \( \text{Area}(S) \) is the area of S. This sum provides a lower bound for the integral, ensuring no overestimation of the area under f. Similarly, for the functions f and g, we can sum their lower sums: \( L(f, P) + L(g, P) \leq L(f+g, P) \), validating that the sum of integrable functions remains integrable.
Upper Sum
The upper sum is similar to the lower sum but uses the supremum of the function values in each sub-interval. For a partition P of domain A, the upper sum U is given by: \[ U(f, P) = \sum_{S \in P} M_S(f) \cdot \text{Area}(S) \] Here, \( M_S(f) \) is the supremum of f in subrectangle S, ensuring we calculate an upper bound for the integral. The upper sum avoids underestimating the function's area. For the sum of the functions f and g, we get: \( U(f+g, P) \leq U(f, P) + U(g, P) \), This means combining integrable functions keeps the sum integrable and aligns their upper bounds, ensuring precision in integration calculations.
Other exercises in this chapter
Problem 2
3-2. Let \(f: A \rightarrow \mathbf{R}\) be integrable and let \(g=f\) except at finitely many points. Show that \(g\) is integrable and \(\int_{A} f=\int_{A} g
View solution Problem 5
3-5. Let \(f, g: A \rightarrow \mathbf{R}\) be integrable and suppose \(f \leq g\). Show that \(\int_{A} f \leq \int_{A} g\)
View solution Problem 6
3-6. If \(f: A \rightarrow \mathbf{R}\) is integrable, show that \(|f|\) is integrable and \(\left|\int_{A} f\right| \leq\) \(\int_{A}|f| .\)
View solution Problem 7
3-7. Let \(f:[0,1] \times[0,1] \rightarrow \mathbf{R}\) be defined by $$ f(x, y)= \begin{cases}0 & x \text { irrational } \\ 0 & x \text { rational, } y \text {
View solution