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Chapter 3: 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 \(\int_{A} f=\int_{A} g\).

Short Answer

Expert verified
Since g differs from f only at finitely many points, g is also integrable and \(\int_{A} g = \int_{A} f\).

Step by step solution

01

- Understand Integrability

First, recall what it means for a function to be integrable. A function is integrable over a set if its definite integral over that set is finite. Since f is given to be integrable, \ \(\int_{A} f\) is finite.
02

- Definition of g

The function g is defined to be equal to f except at a finite number of points in A. Hence, g deviates from f only at those finitely many points.
03

- Effect of Finite Differences

Integral is a measure of area, and changing the function at finitely many points does not affect the area under the curve. The sum of differences over finitely many points has measure zero.
04

- Properties of the Integral

Use the fact that the integral of a function over a set is not affected by changes at finitely many points. Hence, given that f is integrable, g must also be integrable over the set A.
05

- Conclusion

Since g differs from f only at finitely many points and the contribution of these points to the integral is zero, \ \(\int_{A} g = \int_{A} f\). Therefore, we can conclude that g is integrable and the integrals are equal.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
A definite integral can be thought of as the net area under the curve of a function within a specific interval or set. When we calculate the definite integral of a function, we are looking for the total accumulation of the function's value over the interval. In mathematical terms, for a function \(f\) over a set \(A\), the definite integral is expressed as \left( \int_{A} f \right)\. This integral is finite for an integrable function. In our step-by-step problem, knowing \(f\) is integrable tells us that the total area under \(f\) over the set \(A\) is a limited, well-defined value. This lays the foundation for further manipulations and comparisons involving the function \(g\) defined next.
Properties of Integrals
Integrals have useful properties that ease the integration process and help us understand functions better. One crucial property is that changing the values of the function at finitely many points does not alter the integral. This property can be particularly useful when dealing with functions that differ only slightly.
So, for functions \(f\) and \(g\) where \(g = f\) except at finitely many points, the integral over any set \(A\) remains unchanged. This implies that \left( \int_{A} f \right) = \left( \int_{A} g \right)\. This is because the 'area' under a single point is essentially zero in measure theory. Even if there are finitely many such points, their combined contribution to the integral is negligible.
The exercise uses this property to demonstrate that \(g\) is also integrable and equal in integral to \(f\), due to the integral's invariance under finite point changes.
Measure Theory
Measure theory is a branch of mathematics that provides a rigorous way of understanding 'size' or 'measure' of sets, quite similar to but more general than length, area, and volume. In the context of integrals, measure theory helps us understand why certain changes in the function do not affect its integral.
A key concept here is that points or even countable collections of points have measure zero. This means they do not contribute to the integral's total value. Whenever we say that changing the function value at finitely many points does not affect the integral, we are leveraging this concept from measure theory.
For our exercise, measure theory ensures that switching between \(f\) and \(g\)鈥攕ince their difference is confined to finitely many points鈥攄oes not affect the integral's total 'area,' keeping the integrals equal and verifying the integrability of both functions over the set \(A\).

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Most popular questions from this chapter

3-5. Let \(f, g: A \rightarrow \mathbf{R}\) be integrable and suppose \(f \leq g\). Show that \(\int_{A} f \leq \int_{A} g\)

3-35. \(^{*}(a)\) Let \(g: \mathbf{R}^{n} \rightarrow \mathbf{R}^{n}\) be a linear transformation of one of the following types: $$ \begin{aligned} &\left\\{\begin{array}{l} g\left(e_{i}\right)=e_{i} \\ g\left(e_{j}\right)=a e_{j} \end{array} \quad i \neq j\right. \\ &\left\\{\begin{array}{l} g\left(e_{i}\right)=e_{i} \quad i \neq j \\ g\left(e_{j}\right)=e_{j}+e_{k} \end{array}\right. \\ &\left\\{\begin{array}{l} g\left(e_{k}\right)=e_{k} \quad k \neq i, j \\ g\left(e_{i}\right)=e_{j} \\ g\left(e_{j}\right)=e_{i} \end{array}\right. \end{aligned} $$ If \(U\) is a rectangle, show that the volume of \(g(U)\) is \(|\operatorname{det} g| \cdot v(U)\). (b) Prove that \(|\operatorname{det} g| \cdot v(U)\) is the volume of \(g(U)\) for any linear transformation \(g: \mathbf{R}^{n} \rightarrow \mathbf{R}^{n}, \quad\) Hint \(:\) If det \(g \neq 0\), then \(g\) is the composition of linear transformations of the type considered in (a).

\(3-10 .\) (a) If \(C^{\prime}\) is a set of content 0, show that the boundary of \(C\) has content 0 . (b) Give an example of a bounded set \(C\) of measure 0 such that the boundary of \(C\) does not have measure \(0 .\)

3-25. Use induction on \(n\) to show that \(\left[a_{1}, b_{1}\right] \times \times\left[a_{n}, b_{n}\right]\) is not a set of measure 0 (or content 0\()\) if \(a_{i}

3-23. Let \(C \subset A \times B\) be a set of content \(0 .\) Let \(A^{\prime} \subset A\) be the set of all \(x \in A\) such that \(\\{y \in B:(x, y) \in C\\}\) is not of content \(0 .\) Show that \(A^{\prime}\) is a set of measure 0. Hint \(: \chi C\) is integrable and \(\int_{A \times B \chi C}=\int_{A} \mathfrak{U}=\int_{A} \mathcal{L}\), so \(\int_{A} \mathfrak{U}-\mathcal{L}=0\)
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