Covering of Intervals For a real interval $I=[x-r,x+r]$ and a number $\alpha>0$ we write $\alpha I:=[x-\alpha r,x+\alpha r]$, i.e. $\alpha I$ is the interval $I$ blown up by a factor $\alpha$ around its center.
For example, the Vitali Covering Lemma says that for any finite collection of intervals $([a_j,b_j])_{1\leq j\leq n}$ there is an index set $J\subseteq\{1,\ldots,n\}$ such that $([a_j,b_j])_{j\in J}$ is a subcollection of pairwise disjoint intervals with
$$
\bigcup_{1\leq j\leq n}[a_j,b_j]\subseteq\bigcup_{j\in J}3[a_j,b_j]
$$
Setting: I have two finite collections of intervals $([a_j,b_j])_{1\leq j\leq n}$ and $([c_j,d_j])_{1\leq j\leq n}$ with the following three properties.


*

*$|a_j-c_j|\leq\delta$ and $|b_j-d_j|\leq \delta$ for $1\leq j\leq n$,

*$\sum_{j=1}^n|(a_j-c_j)-(b_j-d_j)|\leq\delta$,

*$\left|\bigcup_{j=1}^n[a_j,b_j]\right|\leq\varepsilon$, where $|A|$ means the Lebesgue measure of a set $A$.


Here, $\delta$ and $\varepsilon$ are two positive "small" numbers. 
Problem: I would like to show (or disprove) that the Lebesgue measure of $\bigcup_{j=1}^n[c_j,d_j]$ is, up to a constant that does not depend on the intervals or $n$, like that of $\bigcup_{j=1}^n[a_j,b_j]$ also small.
My attempt: I could prove that this is true if the $[a_j,b_j]$ are mutually disjoint: Indeed, by Vitali's Covering Lemma there is an index set $J\subseteq\{1,\ldots,n\}$ such that $([c_j,d_j])_{j\in J}$ is a subcollection of pairwise disjoint intervals with
$$
\bigcup_{1\leq j\leq n}[c_j,d_j]\subseteq\bigcup_{j\in J}3[c_j,d_j].
$$
This implies with the help of 2. and 3. that
\begin{align*}
\left|\bigcup_{j=1}^n[c_j,d_j]\right|
&\leq\left|\bigcup_{j\in J}3[c_j,d_j]\right|
 \leq\sum_{j\in J}\Big|3[c_j,d_j]\Big|
  =3\sum_{j\in J}|c_j-d_j| \\
&\leq 3\sum_{j\in J}|(a_j-b_j)-(c_j-d_j)|+3\sum_{j\in J}|a_j-b_j| \\
&\leq 3\delta+3\left|\bigcup_{j=1}^n[a_j,b_j]\right| \\
&\leq 3(\delta+\varepsilon).
\end{align*}
However, I have not used 1. In case that the intervals $[a_j,b_j]$ are not pairwise disjoint, I don't know how to proceed. Any ideas are highly appreciated. Thanks in advance!

EDIT: (Thanks to mathworker21): Vitali's Covering Lemma is not necessary here, we can use the estimate
\begin{align*}
\left|\bigcup_{j=1}^n[c_j,d_j]\right|
&\leq\sum_{j=1}^n|c_j-d_j|
\end{align*}
instead.
 A: Here is a disproof. Take any $\epsilon,\delta > 0$. We construct $([a_j,b_j])_{1 \le j \le n}, ([c_j,d_j])_{1 \le j \le n}$ such that:
(1) $|a_j-c_j| \le \delta$ and $|b_j-d_j| \le \delta$ for each $j$.
(2) $\sum_{j=1}^n |(a_j-c_j)-(b_j-d_j)| = 0$.
(3) $|\cup_{j=1}^n [a_j,b_j]| = \epsilon$.
(4) $|\cup_{j=1}^n [c_j,d_j]| = \sqrt{\epsilon \delta n}$.
Take any $L \in \mathbb{N}$, and let $R = \frac{L\delta}{\epsilon}$ and $n = RL$. (It's not a big deal to assume $R \in \mathbb{N}$). For $0 \le l \le L-1$ and $1 \le r \le R$, let $[a_{lR+r},b_{lR+r}] = [l,l+\frac{\epsilon}{L}]$ and $[c_{lR+r},d_{lR+r}] = [l-\delta+(r-1)\frac{\epsilon}{L},l-\delta+r\frac{\epsilon}{L}]$. 
A: (Disclaimer: this is more of a pedantic comment than an answer!)
Does it change the meaning of the question essentially if one drops
the stipulation that $\delta$ and $\epsilon$ are "small" - I
confess to being unable to understand what that statement means in
the present context - and asks instead:

Do there exist a number $h > 0$, and functions $f, g$ (of one and
  two range-limited real variables, respectively), such that if
  $0 < \epsilon < h$, and condition 3 holds, then for all $\delta$
  such that $0 < \delta < f(\epsilon)$, and all $n$ and
  $a_j, b_j, c_j, d_j$ ($1 \leqslant j \leqslant n$) such that
  conditions 1 and 2 hold, we have
  $\left\lvert\bigcup_{j=1}^n[c_j,d_j]\right\rvert \leqslant
g(\epsilon, \delta)$?

A more demanding (but I think less plausible) reformulation of the
question is as follows:

Do there exist numbers $h, h' > 0$, and a function $g$ (of two
  range-limited real variables), such that if $0 < \epsilon < h$ and
  $0 < \delta < h'$, then for all $n$ and $a_j, b_j, c_j, d_j$
  ($1 \leqslant j \leqslant n$) such that conditions 1 to 3 hold, we
  have $\left|\bigcup_{j=1}^n[c_j,d_j]\right| \leqslant
g(\epsilon, \delta)$?

If the second reformulation accurately represents the meaning of the
question, then a special case of mathworker21's argument still gives
the answer "no".
On the other hand, if the first reformulation is satisfactory, then
mathworker21's construction cannot be applied, and the question
remains open.
[Not so! See the addendum below.]
Disposing of the second reformulation first:
Given $h, h' > 0$, take any $\epsilon$ such
that  $0 < \epsilon < \min\{h, h', 1\}$. Let $\delta = \epsilon$, let
$k$ be a positive integer, and, in mathworker21's construction, let
$L = R = k$, $n = k^2$. Then
$\left\lvert\bigcup_{j=1}^n[c_j,d_j]\right\rvert = k\epsilon$.
Because this measure is not bounded above by any function of
$\epsilon$ and $\delta$, the required function $g$ cannot exist.
$\square$
On the first reformulation (which I think is a more reasonable
interpretation of the question), we may still choose any value of
$\epsilon$ that is "small enough", i.e. $\epsilon < h$, and try
to derive a contradiction; but now we must allow $\delta$ to take
on any "sufficiently small" value, i.e. any value less than some
unspecified $f(\epsilon)$. In these circumstances, I don't see how
mathworker21's construction can be carried out.
(I don't know what the answer is in this case, but it seems best to
check my understanding of the question before possibly barking up
the wrong tree!)

Even if the first reformulation is valid, the same general
construction does still apply. This "answer" is therefore
probably best viewed as a long-winded comment in confirmation of
mathworker21's answer.
Given $h > 0$, and a function $f$, take any $\epsilon$ such that
$0 < \epsilon < \min\{h, 1\}$, and any positive integer $k$ such
that $\frac{\epsilon}{k} < f(\epsilon)$.
Define $\delta = \frac{\epsilon}{k}$, In mathworker21's
construction, let $R = k$, $L = k^2$, $n = k^3$. Then
$\left\lvert\bigcup_{j=1}^n[c_j,d_j]\right\rvert = k\epsilon$, much
as before. Again, the required function $g$ cannot exist.
$\square$
