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Chapter 7: Problem 10

Substitutionsregel. Ist \(\varphi:[\alpha, \beta] \rightarrow \mathbb{R}\) monoton wachsend und absolut stetig, \(\varphi(\alpha)=\) : \(a, \varphi(\beta)=: b\) und \(f \in \mathcal{L}^{1}([a, b])\), so ist \((f \circ \varphi) \cdot \varphi^{\prime} \in \mathcal{L}^{1}([\alpha, \beta])\) und $$ \int_{a}^{b} f(x) d x=\int_{\alpha}^{\beta} f(\varphi(t)) \varphi^{\prime}(t) d t $$

Short Answer

Expert verified
The substitution rule allows for equivalencing the given integrals. Upon substitution, the equality \( \int_{a}^{b} f(x) dx = \int_{\alpha}^{\beta} f(\varphi(t)) \varphi^{\prime}(t) dt \) holds. Further, it can be validated that \( (f \circ \varphi) \cdot \varphi^{\prime} \) is indeed Lebesgue integrable over the interval [\(\alpha\), \(\beta\)], as it is a product of two Lebesgue integrable functions.

Step by step solution

01

Interpret the Problem

Given \( \varphi:[\alpha, \beta] \rightarrow \mathbb{R} \), which is monotonically increasing and absolutely continuous, with \( \varphi(\alpha) = a \) and \( \varphi(\beta) = b \), and a function \( f \) that is Lebesgue integrable on [a, b]. The goal is to prove that \( (f \circ \varphi) \cdot \varphi^{\prime} \) is Lebesgue integrable over [\(\alpha\), \(\beta\)] and that the integral of \( f(x) \) from a to b is equal to the integral of \( f(\varphi(t)) \cdot \varphi^{\prime}(t) \) from \(\alpha\) to \(\beta\).
02

Apply the Substitution Rule

In the second step, substitute \( x = \varphi(t) \). The differential \( dx \) is then \( dx = \varphi^{\prime}(t) dt \) where \( \varphi^{\prime}(t) \) indicates the derivative of \( \varphi \) with respect to \( t \).
03

Substitute in the Integral

Substitute \( x \) and \( dx \) into the integral to obtain \( \int_{a}^{b} f(x) dx = \int_{\alpha}^{\beta} f(\varphi(t)) \varphi^{\prime}(t) dt \), reaffirming the equality given in the question.
04

Validate Lebesgue Integrability

Now it is necessary to demonstrate that \( (f \circ \varphi) \cdot \varphi^{\prime} \) is Lebesgue integrable over [\(\alpha\), \(\beta\)]. Recall that both \( f \) and \( \varphi \) are Lebesgue integrable. The product \( (f \circ \varphi) \cdot \varphi^{\prime} \) will also be Lebesgue integrable, as the product of two Lebesgue integrable functions remains Lebesgue integrable.

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Key Concepts

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

Lebesgue Integrable
When we talk about a function being Lebesgue integrable, we're referring to its suitability for integration using the Lebesgue measure. Unlike the Riemann integral, which relies heavily on the notion of summing rectangular areas, the Lebesgue approach is more flexible in handling functions with discontinuities or erratic behavior.

For a function to be Lebesgue integrable, it essentially needs to be bounded and have a finite measure set on which it is defined. This concept extends the idea of integration to a broader class of functions and is particularly useful in complex real-world applications where functions are not always smooth or continuous.

Improving our understanding of Lebesgue integrability equips us to tackle the more challenging integrals encountered in advanced mathematics and applied sciences.
Monotonically Increasing Function
Imagine climbing a hill, each step you take moves you higher than the last鈥攖hat's similar to the concept of a monotonically increasing function. Formally, such a function never decreases as its input increases. For any two inputs, if the second is larger, the output of the function will be either larger or equal to the output at the first input.

This monotonic characteristic guarantees that the function does not have any abrupt changes in direction and that it can be reliably predictable in terms of its growth. In the context of the substitution rule, a monotonically increasing function ensures the substitution will map the integration limits correctly, preventing any overlaps or redundancies which often complicate an integral's evaluation.
Absolute Continuity
The concept of absolute continuity is a stronger form of continuity which combines uniform continuity with an additional condition related to the function's integration. A function is absolutely continuous on an interval if, for any collection of non-overlapping subintervals, the sum of the lengths of those subintervals can be made as small as desired by choosing the subintervals small enough.

Absolutely continuous functions have a series of useful properties: they are continuous, have derivatives almost everywhere, and their derivatives are integrable. When applying the substitution rule in integration, if the substituting function \( \varphi \) is absolutely continuous, it guarantees that it will navigate through any potential 'rough' patches in the function being integrated, thus allowing the substitution rule to be applicable.
Lebesgue Integration
Delving into the world of integration, Lebesgue integration is a powerful mathematical tool that vastly extends the scope of functions that can be integrated, compared to the more traditional Riemann integration. The defining feature of the Lebesgue integration is its ability to measure the size of the set where the function takes on certain values and uses this to calculate the integral.

Lebesgue integration is particularly adept at dealing with functions that oscillate infinitely or have discontinuities鈥攕ituations where the Riemann approach breaks down. By breaking the domain into measurable subsets and establishing a more general integral definition, Lebesgue integration is essential for modern analysis and its applications in probability theory, quantum mechanics, and various other fields.

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Most popular questions from this chapter

Es seien \(\nu, \rho\) signierte Ma脽e auf \(\mathfrak{A}\). a) F眉r \(A \in \mathfrak{A}\) sind folgende Aussagen 谩quivalent: (i) \(A\) ist eine \(\nu\)-Nullmenge. (ii) \(A\) ist eine \(\nu^{+}-\)und eine \(\nu^{-}\)-Nullmenge. (iii) \(A\) ist eine \(\lfloor\nu \|\)-Nullmenge. b) Folgende Aussagen sind 盲quivalent: (i) \(\nu \perp \rho\); (ii) \(\nu^{+} \perp \rho\) und \(\nu^{-} \perp \rho ;\) (iii) \(\mid \nu\rfloor \perp \rho\) (iv) \(|\nu\|\perp\| \rho|\). Sind zus盲tzlich \(\rho, \nu\) endlich, so sind (i)-(iv) auch 盲quivalent \(\mathrm{zu}\) (v) \(|\nu\|\wedge \mid \rho\|=0\).

Es sei \(I \subset \mathbb{R}\) ein Intervall. a) F眉r jede stetige Funktion \(f: I \rightarrow \mathbb{R}\) sind alle Ableitungszahlen Borel-meBbar. b) F眉r jede monotone Funktion \(f: I \rightarrow \mathbb{R}\) sind alle Ableitungszahlen Borel-meBbar. c) F眉r jede Borel- (bzw. Lebesgue-)me脽bare Funktion \(f: I \rightarrow \mathbb{R}\) sind alle Ableitungszahlen Borel- (bzw. Lebesgue-)me脽bar (SIERPI?SKI [1], S. 452 ff., BANACH [1], S. 58 ff., AuERBACH, Fund. Math. 7, 263 (1925), SAKS [2], S. 113 f.). d) F眉r nicht Lebesgue-mebbares \(f: I \rightarrow \mathbb{R}\) brauchen die Ableitungszahlen nicht Lebesgueme脽bar zu sein, k枚nnen aber durchaus Borel- meBbar sein.

Lemma \(3.1\) gilt sinngem盲脽 f眉r \((p, q)=(\infty, 1)\) und liefert f眉r jedes \(\mathrm{MaB} \mu\) eine normerhaltende Injektion \(\varphi: L^{1} \rightarrow\left(L^{\infty}\right)^{\prime} .\) Die Abbildung \(\varphi\) ist nicht surjektiv, falls \(\mu=\beta^{1} \mid \mathfrak{B}_{[0,1]}^{1} .\) (Hinweise: Der Raum \(C([0,1])\) der auf \([0,1]\) stetigen Funktionen mit der Supremumsnorm kann als abgeschlossener Unterraum von \(L^{\infty}(\mu)\) aufgefa脽t werden. Die Abbildung \(\psi: C([0,1]) \rightarrow\) \(\mathbb{K}, \psi(f):=f(0) \quad(f \in C[0,1])\) ist eine stetige Linearform mit \(\|\psi\|=1 .\) Nach dem Satz von HAHN-B ANACH gestattet \(\psi\) eine stetige Fortsetzung \(\Psi: L^{\infty}(\mu) \rightarrow \mathbb{K}\) mit \(\|\psi\|=\|\Psi\| .\) Warum wird \(\Psi\) nicht durch ein Element von \(L^{1}(\mu)\) dargestellt?)

Es seien \(\left(X_{j}, \mathfrak{A}_{j}, \mu_{j}\right)\) ein \(\sigma\)-endlicher MaBraum und \(\nu_{j}\) ein \(\sigma\)-endliches \(\mathrm{Ma} \beta\) auf \(\mathfrak{A}_{j}\) mit der Lebesgueschen Zerlegung \(\nu_{j}=\rho_{j}+\sigma_{j}, \rho_{j} \ll \mu_{j}, \sigma_{j} \perp \mu_{j} \quad(j=1,2)\). Dann hat \(\nu_{1} \otimes \nu_{2}\) bez. \(\mu_{1} \otimes \mu_{2}\) die Lebesguesche Zerlegung \(\nu_{1} \otimes \nu_{2}=\rho+\sigma\) mit \(\rho:=\rho_{1} \otimes \rho_{2} \ll \mu_{1} \otimes \mu_{2}\) und \(\sigma:=\rho_{1} \otimes \sigma_{2}+\rho_{2} \otimes \sigma_{1}+\sigma_{1} \otimes \sigma_{2} \perp \mu_{1} \otimes \mu_{2} .\) Hat \(\rho_{j}\) die Dichte \(f_{j}: X_{j} \rightarrow \mathbb{R}\) bez. \(\mu_{j}(j=1,2)\), so hat \(\rho_{1} \otimes \rho_{2}\) bez. \(\mu_{1} \otimes \mu_{2}\) die Dichte \(f_{1} \otimes f_{2}\left(x_{1}, x_{2}\right):=f_{1}\left(x_{1}\right) \cdot f_{2}\left(x_{2}\right) \quad\left(x_{j} \in X_{j}, j=1,2\right)\).

Ist \(N \subset \mathbb{R}\) eine \(\lambda\)-Nullmenge, so existiert eine wachsende absolut stetige Funktion \(f\) : \(\mathbb{R} \rightarrow \mathbb{R}\), die in keinem \(x \in N\) differenzierbar ist. (Hinweise: Es seien \(U_{n} \supset N, U_{n}\) offen, \(\lambda\left(U_{n}\right)<2^{-n} \quad(n \geq 1)\) und \(\left.f(x):=\int_{0}^{x} \sum_{n=1}^{\infty} \chi_{U_{n}}(t) d t \quad(x \in \mathbb{R}) .\right)\)
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