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Let $Q(\Bbb R)^\uparrow$ be the set of every countable union of half-open intervals, $\lambda: Q(\Bbb R)^\uparrow \rightarrow [0, \infty]$ the extensional Lebesgue-Pre-measure and $\lambda^*: \mathscr P(\Bbb R) \rightarrow [0, \infty]$, $\lambda^*(X)$ $=$ $\inf\{\lambda(B) : X \subset B, B \in Q(\Bbb R)^\uparrow\}$ be the outer Lebesgue-measure. We define:

$v: \mathscr P(\Bbb R) \rightarrow [0, \infty]$,

$v(X) := \inf \{\lambda(A) : A \subset \Bbb R$ open, $X \subset A\}$.

Show that

$\lambda^*(X) = v(X)$.

My attempt:

Since $A$ is an open set, and since every open set $A \subset \Bbb R$ can be written as a countable union of disjoint, half-open intervals with rational endpoints, it follows that $A \in Q(\Bbb R)^\uparrow$.

First, we prove that

$\lambda^*(X) \le v(X)$:

We are allowed to assume that $\lambda^*(X) = \lambda(M)$ for some $M \in Q(\Bbb R)^\uparrow$. This means that there is no $B \in Q(\Bbb R)^\uparrow$ such that $\lambda(B) \lt \lambda(M)$ with $B \supset X$. Therefore, it follows that for every open set $A \in Q(\Bbb R)$, we have that $\lambda(M) \le \lambda(A)$, and therefore, by defintion, $\lambda^*(X) \le v(X)$.

$\lambda^*(X) \ge v(X)$ can be proven analogous, which proves the identity.

Now, since this seems way too easy for me, I guess I made a mistake at some point. Yet, I used the fact that $A$ is open only in order to conclude that it is a part of $Q(\Bbb R)^\uparrow$, but there should be more into it.

Edit:

Asked other people, they couldn't find a mistake, and there doesn't seem to be such a stackexchange-user either. But the more I think about it, the worse does the proof look for me. I simply don't want to believe that the core of the proof is to use that $A \in Q(\Bbb R)^\uparrow$.

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First thing to note : As you said, every open set (except maybe the empty set, but this is not important) is in $Q(\mathbf R)^\uparrow$ so

$$\{\lambda(A): A \text{ is open}, X\subset A\} \subset \{\lambda(B): B\in Q(\mathbf R)^\uparrow , X\subset B\}$$

And taking the inf on both sides gives you automatically $\lambda^*\geq v$, this is the easy part.

The error in your proof is right at the beginning, you cannot assume $\lambda^*(X)=\lambda(M)$ for some $M\in Q(\mathbf R)^\uparrow$ because this is not true. For example if $X=[a;b]$ then for every $M\in Q(\mathbf R)^\uparrow$ such that $X\subset M$ one have $\lambda(M)>\lambda(X)$ (prove it).

Now we want to show that $\lambda^* \leq v$, take a non-empty $X\subset \mathbf R$ and a set $B\in Q(\mathbf R)^\uparrow$ such that $X\subset B$. Now $B= \bigcup_{i\in \mathbf N} [a_i;b_i[$, take $\varepsilon>0$ and define the open set $A$ by

$$A=\bigcup_{i\in \mathbf N}]a_i-\frac{\varepsilon}{2^i}; b_i[.$$

$B$ is included in $A$ so obviously $X\subset A$. Moreover $\lambda(A)\leq \lambda (B)+\varepsilon \sum_{i=0}^\infty 2^{-i}=\lambda(B)+2\varepsilon$. This show that $\lambda^*(X)\leq v(X)+2\varepsilon$ (prove it). Since $\varepsilon$ was arbitrary we get that $\lambda^*(X)\leq v(X)$ for every $X$ and thus $\lambda^*\leq v$.

The proof i gave for $\lambda^*\leq v$ is classic in measure theory.

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  • $\begingroup$ Thanks, glad someone finally pointed out the mistake! :-) $\endgroup$ – Julian Nov 20 '16 at 13:09

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