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Outer Content ( Lebesgue Measure )

  • Thread starter Zondrina
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Zondrina
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Homework Statement



Two parts to this question.

a) Suppose ##S## is a bounded set and has an interior point ##Q##, prove that ##C(S) > 0##. ( Hint : Inside any neighbourhood of ##Q##, you can place a square ).

b) For each integer ##n>0##, let ##L_n## be the line segment ##\{ (\frac{1}{n}, y) : 0 ≤ y ≤ \frac{1}{n} ) \}##.

Also, let ##S = \bigcup_{n=1}^{∞} L_n##, prove that ##C(S) = 0##.

Homework Equations



##C(S)## = Outer content = ##inf \{ \sum A_i \} = inf \{ Area(P) \}##.

Where ##A_i## is the area of the sub-rectangle ##R_i##, which comes from a larger rectangle ##R## that encloses ##S## and has been partitioned by ##P##.

The Attempt at a Solution



a) This part is not too bad I think.

Since ##Q## is an interior point of ##S##, ##\exists \delta > 0 \space | \space N_{\delta}(Q) \subseteq S##.

Now, place a square, with center ##Q##, inside of ##N_{\delta}(Q)## and let ##A_{S_Q} = Area(Square_Q)##.

Suppose that we now enclose ##S## in a rectangle ##R##. For any partition ##P## of ##R##, we can attain another partition ##P'## from the square, which yields the following relationship :

##inf \{ \sum A_i \} ≥ inf \{ \sum A_{i}^{'} \} ≥ A_{S_Q} > 0##.

Hence the result is shown, ##C(S) ≥ A_{S_Q} > 0##.

b) This part is a bit more difficult. I was thinking to take a square with area ##δ^2## that will contain all the line segments ##L_n## except a finite amount of them.
 
Last edited:

Answers and Replies

  • #2
Zondrina
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Sorry for the double, but I think I got this now.

Take a square ##S_1## with area ##δ^2## and vertices at ##(0,0), (δ,0), (0,δ), (δ,δ)## which will contain all except a finite amount of the ##L_n##. So ##\forall ε > 0, \exists δ > 0 \space | \space δ^2 < \frac{ε}{2}##.

Let ##L_{nO} > 0## denote the number of line segments outside of this square.

Enclose each of these line segments in a rectangle ##R_i## such that the area of each rectangle ##A_i## is ##\frac{ε}{2 L_{nO}}## and extend out the sides of each rectangle to get a larger rectangle as well as a partition ##P## of that rectangle, which will contain all the lines. Then ##Area(P)## is defined as :

##Area(S_1) + \sum A_i = δ^2 + \sum A_i##

Now there are ##L_{nO}## line segments, so there must be ##L_{nO}## rectangles we used, therefore :

##δ^2 + \sum A_i = δ^2 + L_{nO} \frac{ε}{2 L_{nO}} < \frac{ε}{2} + \frac{ε}{2} = ε##

Hence ##C(S) = inf \{Area(P)\} < ε, \forall ε > 0##

Therefore ##C(S) = 0##

I think this is the idea. If someone could tell me if Ive gone wrong at all it would be great.
 
Last edited:

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