How to show that $(-\infty,c)$ is measurable by caratheodory criterion?

Let, $m$ be the lebesgue outer measure. Let, $E \subset \mathbb{R}$.

$$m(E) \le m(E \cap (-\infty,c)) + m(E \cap [c,\infty)),$$ is true from definition of outer measure, but I'm having trouble showing the reverse inequality.

Where,

$$m(E) = \inf\{\sum^{\infty}_{j=1}|B_j|: E \subseteq \bigcup^{\infty}_{ j=1}B_j, B_j \textrm{ is a open interval} \}$$

Any help?

Thanks.

It is enough to show,

\begin{align} m(E) &\ge m(E \cap (-\infty,c)) + m(E \cap [c,\infty))\end{align}

Let, $E_1= E\cap (-\infty, c) $ and $E_2= E\cap [c, \infty) $

If $m(E) =\infty$, then nothing to prove.

Assume, $m(E) < \infty$

Then given any $\epsilon>0$ , there exists a countable collection of open intervals $(I_n) $ which covers $E$ such that

$$\sum_{n=1}^{\infty}{\ell(I_n) <m(E) + \epsilon}$$

Let, $I'_n =I_n \cap (-\infty , c) $ and $I''_n =I_n \cap [c, \infty )$

Then, $I'_n \cup I''_n =I_n$ and $I'_n \cap I''_n =\emptyset$

Hence, $\ell(I_n) =\ell(I'_n) + \ell(I''_n ) $

Now, \begin{align}E_1 = E\cap (-\infty, c) &\subseteq \cup_{n=1}^{\infty}I_n\cap (-\infty ,c)\\&=\cup_{n=1}^{\infty}{(I_n\cap (-\infty ,c))}\\ &=\cup_{n=1}^{\infty}I'_n \end{align}

Similarly, $E_2 \subseteq \cup_{n=1}^{\infty}I''_n$

Hence, $m(E_1) =m(\cup_{n=1}^{\infty}I'_n) \le \sum_{n=1}^{\infty} m(I'_n) $

Similarly, $m(E_2) =m(\cup_{n=1}^{\infty}I''_n) \le \sum_{n=1}^{\infty} m(I''_n) $

Hence, \begin{align} m(E_1) +m(E_2 ) &\le \sum_{n=1}^{\infty}\{ m(I''_n)+m(I''_n) \}\\ ​&=\sum_{n=1}^{\infty} \ell(I'_n)+\ell(I_n)\\ &=\sum_{n=1}^{\infty} \ell(I_n)\end{align}

\begin{align} m(E_1 +E_2) &\le \sum_{n=1}^{\infty} \ell(I_n)\\ &<m(E) +\epsilon \end{align}

Since, $\epsilon >0 $ arbitry, it follows that $$m(E_1 +E_2)\le m(E)$$

Hence, the proof is complete.