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Given a subset $A \subset \mathbb{R}$ with the length of an open interval $\mu_L(I_k) = b_k -a_k : I \doteq [a_k,b_k]$

The lebesgue measure is defined as $$ \lambda^{\ast} (A) \doteq \inf \Big\{ \sum_{j \geq 1} \mu_L(I_j) : A \subset \bigcup_{j \geq 1} I_j \Big\} $$

Why is it not defined simply as $$ \lambda^{\ast} (A) \doteq \sum_{j \in J} \mu_L(I_j) : J \rightarrow A = \bigcup_{j \in J} I_j $$

Why is it necessary to have the infimum and why is it necessary for it to be a set? I'm still a little unclear as to why we need to take the infimum

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  • $\begingroup$ $\lambda^\star(A)$ should depend only on $A$ and not on the particular choice of a covering $\{I_j\}$. That's why you want to take the optimal one, that is, the one which minimizes $\sum \mu_L(I_j)$. But unfortunately this minimum might not exist, so one takes the infimum. $\endgroup$ – Giuseppe Negro Oct 21 '13 at 23:13
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The intuition for Lebesgue measure is geometric. Given a reasonable set $X\subset\mathbb{R}^n$, how would you approximate its volume? One way is to approximate its volume from the outside as a sum of volumes of cubes. This is the concept of outer measure.

Say in $\mathbb{R}^2$, to get the area of a disc one might first approximate its area by a square containing the disc. But this is clearly not a good approximation. So let's try chopping this square into 16 squares, 4 rows of 4 squares each, and look at just the squares that intersect the disc. The sum of the areas of these squares (which is well defined, by the way, since we know how to get area of squares) is a better approximation to the area of the circle than our first approximation. But even this is still not a great approximation, so we just keep chopping our rectangles finer and finer. Each approximation is smaller than the last, and is a better approximation to the area of the disc. This explains why we would take the infimum.

In $\mathbb{R}^1$, the intuition is the same, except the sets we are trying to measure are one-dimensional and the "rectangles" become "intervals." We know how to get lengths of intervals, so to get "lengths" of measurable sets we approximate them from the outside as sums of lengths of intervals whose union covers the set.

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  • $\begingroup$ Exactly the answer I was looking for. Thank you. $\endgroup$ – Anthony Peter Oct 22 '13 at 5:43
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There are many sets that you can write as a union of intervals only with degenerate intervals ($[a,a]$), and you need uncountably many intervals. Also, your construction would not produce an outer measure, it would not be subadditive.

Since $A = [0,1]\setminus \mathbb{Q}$ contains no non-degenerate interval, your construction would assign it $\lambda^\ast(A) = 0$ (the only possible interpretation of a sum of uncountably many $0$s is $0$). Also, each singleton set would have $\lambda^\ast(\{x\}) = 0$, but then

$$1 = \lambda^\ast([0,1]) = \lambda^\ast\left(A \cup \bigcup_{r\in [0,1]\cap\mathbb{Q}} \{r\}\right) > \lambda^\ast(A) + \sum_{r\in [0,1]\cap\mathbb{Q}} \lambda^\ast(\{r\}) = 0.$$

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  • $\begingroup$ Could you clarify what $[0,1] \setminus \mathbb{Q}$ represents? I've seen the \ notation before, but I forget what it represents. Is it $[0,1] - \mathbb{Q}$? $\endgroup$ – Anthony Peter Oct 22 '13 at 5:15
  • $\begingroup$ Yes, it's the set difference. One tends to stop using the ordinary minus for it when one uses $A - B$ for $\{ a-b : a \in A,\, b\in B\}$ a lot. $\endgroup$ – Daniel Fischer Oct 22 '13 at 9:12

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