Problem 8
Question
Zeigen Sie mit Hilfe des Riemann-Kriteriums, daß die Funktion $$ f(x)=\operatorname{sgn} x=\left\\{\begin{aligned} 1, & x>0 \\ 0, & x=0 \\ -1, & x<0 \end{aligned}\right. $$ integrierbar über \([a, b]\) ist. (sgn ist die Abkürzung für signum \(=\) Vorzeichen).
Step-by-Step Solution
Verified Answer
The function is Riemann integrable over any interval [a, b].
1Step 1: Identify the Interval and Function Properties
The function \( f(x) = \operatorname{sgn} x \) is defined for \( x \in \mathbb{R} \) with values 1 for \( x > 0 \), -1 for \( x < 0 \), and 0 for \( x = 0 \). We need to show that \( f(x) \) is Riemann integrable on the interval \([a, b]\).
2Step 2: Assess Function's Discontinuities
The function \( f(x) = \operatorname{sgn} x \) has a discontinuity only at \( x = 0 \), where the function jumps from -1 to 1. Check if \( 0 \) lies within the interval \([a, b]\). If not, the function is continuous on \([a, b]\).
3Step 3: Apply the Riemann Criterion
The Riemann criterion states that a function is Riemann integrable on an interval if its set of discontinuities has measure zero. In this case, \( f(x) \) only has a single discontinuity at \( x = 0 \). A single point set has measure zero.
4Step 4: Conclude Integrability
Since \( f(x) \) satisfies the Riemann criterion condition (having discontinuities of measure zero), it is Riemann integrable over any interval \([a, b]\), whether \( 0 \) is included or not.
Key Concepts
Signum FunctionDiscontinuityMeasure ZeroIntegration Criteria
Signum Function
The signum function, often abbreviated as "sgn," is a simple yet fascinating mathematical function. It is defined as follows:
- For any positive number, the signum function produces a value of 1.
- For a negative number, it gives a value of -1.
- At zero, the function outputs 0.
Discontinuity
Discontinuity occurs in functions where there is a sudden jump, break, or value that is undefined within a certain point in the domain. For the signum function, discontinuity presents itself at a notable point:At the point where \( x = 0 \), the signum function \( \operatorname{sgn} x \) skips suddenly from -1 to 1, creating a jump discontinuity. Despite being a single point, this discontinuity is significant because it marks a place of non-smooth transition in the function's graph. In the context of integrability, addressing discontinuities is crucial as they indicate spots where typical calculation becomes complex, yet crucially do not necessarily impact integrability over an interval.
Measure Zero
Measure theory helps us understand sizes of different sets in terms of their 'measure,' which is akin to length, area, or volume. A set is considered to have measure zero if, in a sense, it occupies no space within the interval it's located.For the signum function, the discontinuity at\( x = 0 \)gives rise to a set that contains just one point. Since any single point in real numbers has measure zero, the discontinuity \( \{0\} \) does not contribute a meaningful 'size' or 'length' when considering measure. This insight is essential for applying integration criteria, linking the concept of measure with the ability to integrate despite discontinuities.
Integration Criteria
The Riemann integration criteria offers a method to check a function's integrability over an interval. For a function to be Riemann integrable:
- Its discontinuities must have measure zero.
Other exercises in this chapter
Problem 4
Gegeben sei die äquidistante Zerlegung \(Z=\left\\{x_{0}, \ldots, x_{n}\right)\) mit \(\left(x_{1}-x_{i-1}\right)=\frac{b-a}{n}\) für alle \(i \in\\{1, \ldots,
View solution Problem 6
Zeigen Sie die Integrierbarkeit von \(f(x)=c\), \(c \in \mathbb{R}\), über \([a, b] .\) Was ergibt der Vergleich mit der elementaren Definition auf Seite \(94 ?
View solution Problem 11
Zeigen Sie, daB die Funktion \(f(x)=\sqrt{x}\) integrierbar über \([O, b]\) ist und bestimmen Sie mit Satz (7.17) den Flächeninhalt \(\int_{0}^{b} \sqrt{x} d x\
View solution Problem 14
Bestimmen Sie mit der Trapezregel für \(n=2\), 4 und 8 Näherungen von $$ \int_{0}^{1} \frac{1}{2-x^{2}} d x=0,623225 \ldots $$ Wie grós sind die Fehler?
View solution