# Outer Content ( Lebesgue Measure )

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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.

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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.

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