The problem:

Let $E$ be a Lebesgue measurable subset of $\mathbb{R}^d$ with finite measure. To show that $\forall \epsilon>0$, $\exists$ a compact set $V$ such that $m(E \setminus V) \leq \epsilon$.

My approach:

$E$ is Lebesgue measurable. Fix $\epsilon > 0$. Now we can employ the inner approximation by closed set criterion for Lebesgue measurability to assert that $\exists$ a closed set $F$, contained in $E$, such that $m(E \setminus F) \leq \frac{\epsilon}{3}$.

Now if $E$ is bounded, so is $F$ and we are done. So let us assume that $E$ is unbounded, so that $F$ is not necessarily bounded. By monotonicity, $m(F) \leq m(E) < \infty$.

Let us decompose $\mathbb{R}^d$ into countable number of almost disjoint closed bounded boxes. More formally, $\mathbb{R}^d=\bigcup_{n=1}^{\infty}B_n$, where $B_n$$'$s are almost disjoint closed boxes. Now let $G$ be the set of all these closed boxes which are completely contained inside $F$. Formally written, $$G=\bigcup\Big\{B_{n_j} : B_{n_j} = B_n ~\mbox{for some}~ n \in \mathbb{N} ~\mbox{and}~ B_{n_j} \subset F\Big\}$$ Clearly $G \subset F$. Since $G$ is countable union of almost disjoint closed boxes, $m(G)=\sum_{j=1}^{\infty}|B_{n_j}|$. Again by monotonicity, $m(G) \leq m(F) < \infty$. Hence the sum $\sum_{j=1}^{\infty}|B_{n_j}|$ converges. Now we can choose $N \in \mathbb{N}$, which depends on the pre-fixed $\epsilon$, such that the partial sum $\sum_{j=1}^N|B_{n_j}| \geq \sum_{j=1}^{\infty}|B_{n_j}|-\frac{\epsilon}{3}$.

Define $S_N=\bigcup_{n_j=1}^N B_{n_j}$. Clearly, $S_N \subset G$. $S_N$ is closed and bounded (being finite union of closed, bounded boxes) and hence compact, by the Heine-Borel theorem. Thus it is a possible candidate for the required compact set contained in $E$.

Now, by additivity of Lebesgue measure, $m(G)=m(S_N)+m(G \setminus S_N)$. Then

$m(G \setminus S_N)=m(G)-m(S_N)=\sum_{j=1}^{\infty}|B_{n_j}| - \sum_{j=1}^N|B_{n_j}| \leq \frac{\epsilon}{3}$.

Finally we need to control the quantity $m(E \setminus S_N)$. Note that $$E \setminus S_N = (E \setminus F) ~\bigcup~ (F \setminus G) ~\bigcup~ (G \setminus S_N)$$ Each of the sets in the right-hand side of the above equation is Lebesgue measurable (each being intersection of two Lebesgue measurable sets). Using additivity, we have $$m(E \setminus S_N)=m(E \setminus F)+m(F \setminus G)+m(G \setminus S_N)$$

We have already shown that $m(E \setminus F) \leq \frac{\epsilon}{3}$ and $m(G \setminus S_N) \leq \frac{\epsilon}{3}$.

All that is left is to control the quantity $m(F \setminus G)$. It is intuitively very clear that we should be able to keep this quantity as small as possible, whenever $E$ is Lebesgue measurable, making the decomposition of $\mathbb{R}^d$ increasingly finer. What I need is a rigourous argument to defend this point. Any help would be greatly appreciated.

EDIT: I think that I found a logic. Suppose $m(F \setminus G) > \frac{\epsilon}{3}$. Since $F \setminus G$ is Lebesgue measurable, we can use the inner approximation by closed criterion for Lebesgue measurability to assert that $\exists$ a closed set $H$, contained in $F \setminus G$ such that $m((F \setminus G) \setminus H) \leq \frac{\epsilon}{3}$. Then we can include $H$ to $G$ and replace $G$ with $G^*=G \cup H$ in the above argument and continue. Please check if this reasoning is good enough.

  • $\begingroup$ In general, you can't approximate $F$ from the 'inside' with boxes with have non empty interiors. Take the Cantor set as an example. It has an empty interior. $\endgroup$ – copper.hat Dec 17 '16 at 18:57

You are doing too much work.

Suppose you have $F\subset E$ closed such that $m (E \setminus F) < \epsilon$.

If $F$ is bounded then you are finished (since then $F$ is compact), so all we need to deal with is the case where $F$ might be unbounded.

Consider $F_n = F \cap [-n,n]^d \subset F$. Note that $F_n \subset F_{n+1}$ and each $F_n$ is closed and bounded hence compact.

Furthermore, $\cup_n F_n = F$ and so $E \setminus F = \cap_n (E \setminus F_n)$.

Since $E \setminus F_n \subset E$, we can use continuity of measure to get $m(E \setminus F_n) \to m (E \setminus F)$ and hence for $n$ large enough, we have $m (E \setminus F_n) < \epsilon$. (Note how we used the fact that $mE < \infty$ to ensure convergence of measure.)

To see why finite measure is necessary, note that for any compact $K \subset \mathbb{R}^d$ we must have $m (\mathbb{R}^d \setminus K) = \infty$.


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