Question

(a) Suppose that \(f\) is integrable on \(S\) and \(S_{0}\) is obtained by removing a set of zero content from \(S\). Show that \(f\) is integrable on \(S_{0}\) and \(\int_{S_{0}} f(\mathbf{X}) d \mathbf{X}=\) \(\int_{S} f(\mathbf{X}) d \mathbf{X}\) (b) Prove Corollary 7.1.31.

Short answer

Answer: Yes, if a function f is integrable on a region S and S_1 is a subset of zero content in S, then f is also integrable on S \ S_1, and the integrals of f over both regions are equal.

Step-by-step solution

We're given that \(f\) is integrable on \(S\). We need to show that \(f\) is integrable on \(S_0\) (which is obtained by removing a set of zero content from \(S\)) and that the integral of \(f\) over both regions is the same. #Step 2: Proving f is integrable on S_0#

Since \(f\) is integrable on \(S\), for any \(\varepsilon > 0\), there exists a partition \(P\) of \(S\) such that the upper sum \(U(f,P)\) and lower sum \(L(f,P)\) satisfy: \(U(f,P) - L(f,P) < \varepsilon\) Define a partition \(P_0\) for \(S_0\). Since the set of zero content removed is a subset of \(S\), we have \(U(f,P_0) \leq U(f,P)\) and \(L(f,P_0) \geq L(f,P)\). Hence, \(U(f,P_0) - L(f,P_0) \leq U(f,P) - L(f,P) < \varepsilon\) Since this holds for any \(\varepsilon > 0\), \(f\) is integrable on \(S_0\). #Step 3: Proving the integrals over S_0 and S are equal#

We have already proved that \(f\) is integrable on \(S_0\). Now we need to show that the integral of \(f\) over \(S_0\) is the same as the integral of \(f\) over \(S\). Since the set removed has zero content, we can write: \(\int_{S} f(\mathbf{X}) d \mathbf{X} - \int_{S_0} f(\mathbf{X}) d \mathbf{X} = \int_{S \setminus S_0} f(\mathbf{X}) d \mathbf{X} = 0\) Hence, \(\int_{S_0} f(\mathbf{X}) d \mathbf{X} = \int_{S} f(\mathbf{X}) d \mathbf{X}\) Now, let's move on to part (b). #Step 4: Proving Corollary 7.1.31#

The corollary states that if \(f\) is integrable on \(S\) and \(S_1\) is a subset of zero content in \(S\), then \(f\) is integrable on \(S \setminus S_1\) and \(\int_{S \setminus S_1} f(\mathbf{X}) d \mathbf{X} = \int_{S} f(\mathbf{X}) d \mathbf{X}\). Let \(S_0 = S \setminus S_1\), where \(S_1\) is a subset of zero content in \(S\). Using the result from part (a), we've already shown that if \(f\) is integrable on \(S\), then it is also integrable on \(S_0\) and that the integrals over both regions are the same. Thus, the corollary is proven, i.e., if \(f\) is integrable on \(S\) and \(S_1\) is a subset of zero content in \(S\), then \(f\) is integrable on \(S \setminus S_1\) and \(\int_{S \setminus S_1} f(\mathbf{X}) d \mathbf{X} = \int_{S} f(\mathbf{X}) d \mathbf{X}\).

Key concepts

Zero Content Sets

In the realm of mathematics, particularly in measure theory, the concept of zero content sets is fundamental. A set is said to have zero content if its size, measured by its length, area, or volume, is effectively zero. This means we can think of these sets as being so "thin" that they don't contribute anything substantial to an integral. When we integrate a function over a region, removing a zero content set from that region leaves the integral unchanged.
This is because these sets do not contain enough "space" to alter the value of the integral.

• Zero content often involves boundaries or points without dimension, such as a surface in a 3D space having area zero, or a line in a 2D plane having length zero.
• The idea is crucial when simplifying and adjusting the regions over which we integrate, particularly ensuring that even when a region is modified, the result remains consistent.

Set Difference in Integration

The set difference in integration is an insightful concept when dealing with integrals on modified regions. Essentially, when you take a set, say \(S\), and remove another set, \(S_0\), the remaining part, \(S \setminus S_0\), is the set difference. In integration, the focus is on whether this difference affects the integral's value.

• As seen in the exercise, if \(S_0\) has zero content, the integral over \(S\) and \(S \setminus S_0\) will be identical.
• This results from the fact that the function's contribution on the zero content set is negligible.

Understanding this helps in refining regions without altering integral results, a key technique in advanced calculus and analysis.

Integrable Functions

An integrable function is one that can be integrated over a given region, meaning its integral exists and yields a finite number. For Lebesgue integration, functions are judged based on how well they can be approximated by simple functions under a measure, allowing this powerful integration method to account for more complex functions compared to Riemann integration.

• A function \(f\) is integrable over a set if for every positive \(\varepsilon\), there exist partitions such that the difference between the upper and lower sums is less than \(\varepsilon\).
• This ensures the integral is well-defined, even if the function has discontinuities.
• Integrable functions allow the use of the integral in solving real-world problems and proving mathematical theorems.

When zero content sets are involved, as long as the original function is integrable, the removal of these sets doesn't affect integrability or the value of the integral.