Question

Let there exist \(\int_{a}^{b}|f(x)| d x .\) Does it follow that the function \(\mathrm{f}(\mathrm{x})\) is integrable on the closed interval \([a, b] ?\)

Short answer

Explain why or why not. Answer: Yes, if the absolute value of a function f(x) is integrable on a closed interval [a, b], it guarantees that the function f(x) is also integrable on the same interval. This conclusion can be drawn by investigating the relationship between the upper and lower sums of f(x) and |f(x)|, and showing that the difference between the upper and lower sums of f(x) converges to zero as the partition of the interval becomes finer.

Step-by-step solution

Recalling the definition of integrability

A function f(x) is said to be Riemann integrable on the closed interval [a, b] if the upper and lower sums of the function approach the same limit as the partition of the interval becomes finer and finer. In other words, if \(\lim_{n\to\infty} \left( U_n(f) - L_n(f) \right) = 0\). Now let's consider how this definition applies to both f(x) and |f(x)|. #Phase2#

Integrability of |f(x)|

Since we know that |f(x)| is integrable on the interval [a, b], it follows that the upper and lower sums of |f(x)| approach the same limit as the partition of the interval becomes finer. This can be written as \(\lim_{n\to\infty} \left( U_n(\|f\|) - L_n(\|f\|) \right) = 0\).

Comparing the integrability of f(x) and |f(x)|

To determine if f(x) is integrable, we need to examine the relationship between the upper and lower sums of f(x) and those of |f(x)|. For any partition P, we shall denote \(M_i\) and \(m_i\) as the supremum and infimum of f(x) inside an interval \([x_{i-1},x_i]\), respectively. Note that: 1) \(|M_i|\leq \max\{|m_i|,|M_i|\}=M_i(\|f\|)\). 2) \(|m_i|\leq \max\{|m_i|, |M_i|\}=M_i(\|f\|)\). 3) \(M_i(\|f\|)-m_i(\|f\|)\geq M_i+m_i\). Since |f(x)| is integrable, it follows that \(\lim_{n\to\infty} \left( U_n(\|f\|) - L_n(\|f\|) \right) = 0\). We can write the same inequality for f(x) as follows: \(\lim_{n\to\infty}\left(\sum_{i=1}^n|M_i\Delta x_i| - \sum_{i=1}^n|m_i\Delta x_i|\right) \leq \lim_{n\to\infty} (U_n(\|f\|) - L_n(\|f\|))=0\). This inequality tells us that the difference between the upper and lower sums of f(x) converges to zero as n goes to infinity. As a result, f(x) is indeed integrable on the closed interval [a, b].

Key concepts

Riemann Integrable

Integrability is a cornerstone in the world of calculus, especially when dealing with the Riemann integral. A function is considered Riemann integrable if it satisfies a specific set of conditions over a closed interval, typically noted as \[a, b\]. For a function \(f(x)\), its integrability is determined by examining the behavior of its upper and lower sums, \(U_n(f)\) and \(L_n(f)\), as we refine the partition of the interval into smaller subintervals.

The core idea is to look at whether these sums converge to the same value, in the limit, as the number of subintervals increases indefinitely. Mathematically, we say \(f(x)\) is Riemann integrable if: \[ \lim_{n\to\infty} \left( U_n(f) - L_n(f) \right) = 0 \]. The convergence of the upper and lower sums to zero implies that the 'gaps' in our approximation of the area under \(f(x)\) get infinitesimally small, allowing us to define the definite integral of \(f(x)\).

Upper and Lower Sums

Delving deeper into the concept of integrability, we encounter the upper and lower sums. These are methods to approximate the area under the curve of a function by creating a series of rectangles over the interval \[a, b\].

For upper sums, \(U_n(f)\), we take the supremum (the least upper bound) of \(f(x)\) over each partition interval and use it to calculate the area of the rectangle. Conversely, for lower sums, \(L_n(f)\), we use the infimum (the greatest lower bound) of \(f(x)\) over each partition interval. In a graphical sense, upper sums overestimate while lower sums underestimate the area under \(f(x)\). The key is that as we make more partitions, the maximum error between the actual area and our approximation diminishes, and we aim for these sums to become arbitrarily close to each other and to the true area.

Absolute Values in Integration

Considering the absolute value of a function, \( |f(x)| \) within the realm of integration offers an interesting aspect. When we integrate the absolute value of a function, we're looking at the area above and below the x-axis as always positive.

The presence of absolute value affects the integrability of \( f(x) \) in an impactful way. The exercise suggests that the fact \( \int_{a}^{b}|f(x)| \,dx \) exists does not immediately guarantee the integrability of \( f(x) \) without absolute values. However, through the logic of upper and lower sums, the integrability of \( |f(x)| \) can imply the integrability of \( f(x) \) under certain conditions. As step 3 of the solution outlines, since the absolute values ensure non-negative differences, the sums involved obey particular inequalities that ultimately lead \( f(x) \) to be Riemann integrable if its absolute version is.

Limit of a Sequence

The limit of a sequence is a fundamental concept in calculus when we speak about convergence. Consider a sequence of numbers that we can imagine as a list of outcomes from some process that's performed over and over. The limit is what these outcomes tend closer and closer to as we repeat the process more and more.

In the context of Riemann integrability, we look at the limit of sequences formed by the upper and lower sums, \(U_n(f)\) and \(L_n(f)\), as the number of partitions (denoted by \(n\)) goes to infinity. If these two sequences converge to the same number, that suggests our method of approximating the area under \(f(x)\) becomes very precise in the limit, hence solidifying the notion that the limit of a sequence is pivotal to the concept of integrable functions.