How Does Function Behavior in Intersection Impact Integrals in Rectangles?

In summary, the theorem states that if two rectangles Q and Q' in R^n have a bounded function F: R^n -> R that is 0 outside their intersection, then the integral of F over Q is equal to the integral of F over Q'. This is because the function F is only determined by its values in the intersection, not outside of it. Additionally, if R is a subrectangle that is not contained in the intersection, then F vanishes at some points in R. However, if R is a subset of the intersection, then F is strictly equal to 0 on R. This is because the integral of F is determined by the values in the intersection. The theorem also states that the lower sum of F over a
  • #1
Buri
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I don't fully understand this theorem:

Let Q and Q' be two rectangles in R^n. If F: R^n -> R is a bounded function that vanishes outside Q intersect Q', then integral of f over q is equal to the integral of f over Q'.

When it says that the function vanishes outside of Q intersect Q', does it mean its always zero in the intersection, or simply at some points? See if it only vanishes at some points I don't see how the theorem could be true. The proof considers the special case when Q is a subset of Q' and the partitions Q'. Then it creates a refinement of that partition by adjoining the endpoint of Q into the partition of Q'. However, it goes on to say that if R is a sub rectangle that is not contained in Q then f vanishes at SOME point in R. But I understood that it vanishes in the intersection and thus should completely vanishes on R.

What am I not understanding?
 
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  • #2
f is 0 outside of the intersection. Therefore the integral of f is determined by its value in the intersection. Q = Q∩Q' + Q-Q' and Q'=Q∩Q' + Q'-Q. Since f = 0 on both Q-Q' and Q'-Q, the integral of f is the integral over Q∩Q', which means the integral over Q and the integral over Q' are the same.
 
  • #3
Why does it say then that "then f vanishes at SOME point in R" and not rather in ALL of R?

Using what it says it says that it follows that m_R(f) is less than or equal to 0. But if f(x) were equal to 0 on all of R which it seems like it should be (since R isn't in the intersection but rather only in Q'), then m_R (f) shouldn't be less than zero.

It goes on to say that:

L(f,P") less than or equal to sum of [m_R(f)]vol(R), where R are the sub rectangles contained in Q and m_R (f) = inf{f(x)| x in R}.

Thanks for your help, mathman I appreciate it.
 
  • #4
If R is not a subset of the intersection, then some points of R are in the intersection (maybe none of them but we don't know), and some points are outside the intersection
 
  • #5
Ahh yes that's true. I guess the only R which may have m_R (f) < 0 are the ones which are along the boundary of Q as they have some points of Q inside. If they don't, however, m_R (f) should be strictly equal to 0.

Thanks a lot for your help Office Shredder! :)
 
  • #6
Let Q and Q' be two rectangles in R^n. If F: R^n -> R is a bounded function that vanishes outside Q intersect Q', then integral of f over q is equal to the integral of f over Q'.

That was the part of your comment I was trying to address. Your further comment seems to be a distortion of the original problem.

When it says that the function vanishes outside of Q intersect Q', does it mean its always zero in the intersection, or simply at some points?
This statement is completely wrong! The original statement is the function is 0 OUTSIDE the intersection, and anything INSIDE!
 

FAQ: How Does Function Behavior in Intersection Impact Integrals in Rectangles?

What is an "integral over a bounded set"?

An integral over a bounded set refers to the process of finding the total value of a function within a specific bounded region or set of values. This is often represented as the area under the curve of a function within a certain range of inputs.

How is an integral over a bounded set calculated?

To calculate an integral over a bounded set, you first need to determine the bounds of the set. This can be done by finding the minimum and maximum values of the independent variable within the set. Then, the integral is evaluated using a specific method, such as the Riemann sum or the Fundamental Theorem of Calculus.

What is the significance of an integral over a bounded set?

An integral over a bounded set has many applications in mathematics, physics, and engineering. It can be used to find the area under a curve, calculate the volume of a solid, and solve optimization problems. It is also essential in understanding the concept of accumulation, as the integral represents the accumulation of a function over a given range.

Can an integral over a bounded set be negative?

Yes, an integral over a bounded set can be negative if the function being integrated has negative values within the set. This can happen if the curve of the function dips below the x-axis, resulting in a negative area under the curve. In this case, the integral represents the net area between the curve and the x-axis.

How is an integral over a bounded set different from an indefinite integral?

An integral over a bounded set has specific limits of integration, while an indefinite integral does not. This means that an indefinite integral gives a general solution, while an integral over a bounded set gives a particular value. Additionally, the integral over a bounded set is a finite number, while an indefinite integral is represented by a constant and a variable.

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