We prove that every Borel set is Lebesgue measurable, establishing that $\mathcal{B}(\mathbb{R}) \subseteq \mathcal{L}$.

Proof: By the Caratheodory criterion, we need to show that for every open interval $I = (a, b)$ and every set $A \subseteq \mathbb{R}$: $\lambda^*(A) \geq \lambda^*(A \cap I) + \lambda^*(A \cap I^c)$

The reverse inequality follows from subadditivity of outer measure, so equality holds if we establish this inequality.

Step 1: We may assume $\lambda^*(A) < \infty$, otherwise the inequality is trivial.

Step 2: Let $\epsilon > 0$. By definition of outer measure, there exists a countable collection of intervals $\{J_k\}$ covering $A$ such that: $\sum_{k=1}^{\infty} \ell(J_k) < \lambda^*(A) + \epsilon$

Step 3: For each $k$, we can write: $J_k = (J_k \cap I) \cup (J_k \cap I^c)$

Using the subadditivity property for lengths: $\ell(J_k) \geq \ell(J_k \cap I) + \ell(J_k \cap I^c)$

(This follows because $J_k \cap I$ and $J_k \cap I^c$ are disjoint, and their union covers $J_k$ up to at most two boundary points.)

Step 4: Therefore: $\sum_{k=1}^{\infty} \ell(J_k) \geq \sum_{k=1}^{\infty} \ell(J_k \cap I) + \sum_{k=1}^{\infty} \ell(J_k \cap I^c)$

Step 5: Since $\{J_k \cap I\}$ covers $A \cap I$ and $\{J_k \cap I^c\}$ covers $A \cap I^c$: $\sum_{k=1}^{\infty} \ell(J_k) \geq \lambda^*(A \cap I) + \lambda^*(A \cap I^c)$

Step 6: Combining with Step 2: $\lambda^*(A) + \epsilon > \lambda^*(A \cap I) + \lambda^*(A \cap I^c)$

Since $\epsilon$ was arbitrary, $\lambda^*(A) \geq \lambda^*(A \cap I) + \lambda^*(A \cap I^c)$.