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- I am reading Multidimensional Real Analysis II (Integration) by J.J. Duistermaat and J.A.C. Kolk ... and am focused on Chapter 6: Integration ...

I need some help with aspects of the proof of Proposition 6.1.2 ...

I am reading Multidimensional Real Analysis II (Integration) by J.J. Duistermaat and J.A.C. Kolk ... and am focused on Chapter 6: Integration ...

I need some help with the proof of Proposition 6.1.2 ...

Proposition 6.1.2 reads as follows:

Definitions and text preliminary to the Proposition reads as follows:

I am trying to write a detailed proof of the fact or assertion that

[itex] \text{vol}_n (B) = \text{vol}_n (B') + \text{vol}_n (B'') [/itex] ... ... ... ... (*)

given their definitions as sets ... but have so far been able to formulate a proof ...

I hope someone can help ...

I am especially interested in tying or connecting the proof of (*) to the definitions of B' and B" as sets.

To explicitly indicate my concerns I am presenting an example from $$ \mathbb{R^2} $$

Let [itex] B' = \{ ( x_1, x_2 ) \in \mathbb{R^2} \; \; | \; \; a_1 \leq x_1 \leq t_1 \text{ and } a_2 \leq x_2 \leq b_2 \; \; ; \; \; \text{ and } a_2 \leq x_2 \leq t_2 \text{ and } a_1 \leq x_1 \leq b_1 \} [/itex]

Thus B' is made up of two rectangles, viz.

[itex] B'_1 = \{ ( x_1, x_2 ) \in \mathbb{R^2} \; \; | \; \; a_1 \leq x_1 \leq t_1 \text{ and } a_2 \leq x_2 \leq b_2 \} [/itex]

and

[itex] B'_2 = \{ ( x_1, x_2 ) \in \mathbb{R^2} \; \; | \; \; a_2 \leq x_2 \leq t_2 \text{ and } a_1 \leq x_1 \leq b_1 \} [/itex]

These two rectangles are depicted in Figures 1 and 2 below ...

Now ... let [itex] B" = \{ ( x_1, x_2 ) \in \mathbb{R^2} \; \; | \; \; t_1 \leq x_1 \leq b_1 \text{ and } a_2 \leq x_2 \leq b_2 \; \; ; \; \; \text{ and } t_2 \leq x_2 \leq b_2 \text{ and } a_1 \leq x_1 \leq b_1 \} [/itex]

Thus [itex] B" [/itex] is made up of two rectangles, viz.

[itex] B"_1 = \{ ( x_1, x_2 ) \in \mathbb{R^2} \; \; | \; \; t_1 \leq x_1 \leq b_1 \text{ and } a_2 \leq x_2 \leq b_2 \} [/itex]

and

[itex] B"_2 = \{ ( x_1, x_2 ) \in \mathbb{R^2} \; \; | \; \; t_2 \leq x_2 \leq b_2 \text{ and } a_1 \leq x_1 \leq b_1 \} [/itex]

The above two rectangles [itex] B"_1 [/itex] and [itex] B"_2 [/itex] are depicted in Figures 3 and 4 below ...

Now ... my problem is this: I do not see how

[itex] \text vol_n (B) =\text vol_n (B') + \text vol_n (B'') [/itex] ... ... ... ... (*)

... that is I do not see how the volumes of [itex] B' [/itex] and [itex] B" [/itex] which together comprise the overlapping rectangles [itex] B'_1 , B'_2 , B"_1 \text{ and } B"_2 [/itex] satisfy (*) ...

Can someone explain how (*) can be true given my analysis above ... I must be making some errors ...

NOTE: I know that (6.2) in the text in defining the conditions for a partition of a rectangle Duistermaat and Kolk write

[itex] B_i \cap B_j = \emptyset [/itex] or [itex] \text vol_n ( B_i \cap B_j ) = 0 [/itex] if [itex] i \neq j [/itex]

... but this is prescriptively defining a partition of a rectangle [itex] B [/itex]... it does not, in my opinion, mean that we can overlook (put to zero) the overlaps or intersections in the rectangles [itex] B'_1 , B'_2, B"_1, \text{ and } B"_2 [/itex] ...

Either way I cannot see how [itex] \text vol_n (B) =\text vol_n (B') + \text vol_n (B'') [/itex] ... ... ... ... (*)

arises from the union of the rectangles

[itex] B'_1 , B'_2, B"_1, \text{ and } B"_2 [/itex] ...

Hope someone can explain and clarify this aspect of the proof of Proposition 6.1.2 ...

Peter

I need some help with the proof of Proposition 6.1.2 ...

Proposition 6.1.2 reads as follows:

Definitions and text preliminary to the Proposition reads as follows:

I am trying to write a detailed proof of the fact or assertion that

[itex] \text{vol}_n (B) = \text{vol}_n (B') + \text{vol}_n (B'') [/itex] ... ... ... ... (*)

given their definitions as sets ... but have so far been able to formulate a proof ...

I hope someone can help ...

I am especially interested in tying or connecting the proof of (*) to the definitions of B' and B" as sets.

To explicitly indicate my concerns I am presenting an example from $$ \mathbb{R^2} $$

Let [itex] B' = \{ ( x_1, x_2 ) \in \mathbb{R^2} \; \; | \; \; a_1 \leq x_1 \leq t_1 \text{ and } a_2 \leq x_2 \leq b_2 \; \; ; \; \; \text{ and } a_2 \leq x_2 \leq t_2 \text{ and } a_1 \leq x_1 \leq b_1 \} [/itex]

Thus B' is made up of two rectangles, viz.

[itex] B'_1 = \{ ( x_1, x_2 ) \in \mathbb{R^2} \; \; | \; \; a_1 \leq x_1 \leq t_1 \text{ and } a_2 \leq x_2 \leq b_2 \} [/itex]

and

[itex] B'_2 = \{ ( x_1, x_2 ) \in \mathbb{R^2} \; \; | \; \; a_2 \leq x_2 \leq t_2 \text{ and } a_1 \leq x_1 \leq b_1 \} [/itex]

These two rectangles are depicted in Figures 1 and 2 below ...

Now ... let [itex] B" = \{ ( x_1, x_2 ) \in \mathbb{R^2} \; \; | \; \; t_1 \leq x_1 \leq b_1 \text{ and } a_2 \leq x_2 \leq b_2 \; \; ; \; \; \text{ and } t_2 \leq x_2 \leq b_2 \text{ and } a_1 \leq x_1 \leq b_1 \} [/itex]

Thus [itex] B" [/itex] is made up of two rectangles, viz.

[itex] B"_1 = \{ ( x_1, x_2 ) \in \mathbb{R^2} \; \; | \; \; t_1 \leq x_1 \leq b_1 \text{ and } a_2 \leq x_2 \leq b_2 \} [/itex]

and

[itex] B"_2 = \{ ( x_1, x_2 ) \in \mathbb{R^2} \; \; | \; \; t_2 \leq x_2 \leq b_2 \text{ and } a_1 \leq x_1 \leq b_1 \} [/itex]

The above two rectangles [itex] B"_1 [/itex] and [itex] B"_2 [/itex] are depicted in Figures 3 and 4 below ...

Now ... my problem is this: I do not see how

[itex] \text vol_n (B) =\text vol_n (B') + \text vol_n (B'') [/itex] ... ... ... ... (*)

... that is I do not see how the volumes of [itex] B' [/itex] and [itex] B" [/itex] which together comprise the overlapping rectangles [itex] B'_1 , B'_2 , B"_1 \text{ and } B"_2 [/itex] satisfy (*) ...

Can someone explain how (*) can be true given my analysis above ... I must be making some errors ...

NOTE: I know that (6.2) in the text in defining the conditions for a partition of a rectangle Duistermaat and Kolk write

[itex] B_i \cap B_j = \emptyset [/itex] or [itex] \text vol_n ( B_i \cap B_j ) = 0 [/itex] if [itex] i \neq j [/itex]

... but this is prescriptively defining a partition of a rectangle [itex] B [/itex]... it does not, in my opinion, mean that we can overlook (put to zero) the overlaps or intersections in the rectangles [itex] B'_1 , B'_2, B"_1, \text{ and } B"_2 [/itex] ...

Either way I cannot see how [itex] \text vol_n (B) =\text vol_n (B') + \text vol_n (B'') [/itex] ... ... ... ... (*)

arises from the union of the rectangles

[itex] B'_1 , B'_2, B"_1, \text{ and } B"_2 [/itex] ...

Hope someone can explain and clarify this aspect of the proof of Proposition 6.1.2 ...

Peter

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