I am teaching myself measure theory over the summer, and would appreciate some feedback on this basic result, since this type of reasoning is quite new to me.
Solution:
By Heine-Borel, the compact sets of $\mathbb{R}$ are exactly the closed bounded subsets. Let $\mathcal{K}(\mathbb{R})$ denote the $\sigma$ algebra generated by the compact sets. The Borel $\sigma$ algebra on $\mathbb{R}$, $\mathcal{B}(\mathbb{R})$, is generated by the open subsets of $\mathbb{R}$, and therefore the closed subsets of $\mathbb{R}$. Additionally, $\mathcal{B}(\mathbb{R})$ is generated by all the sets of the form $(-\infty,t]$ for real $t$. Clearly, the compact sets are contained in the collection of closed sets, so $\mathcal{K}(\mathbb{R}) \subset \mathcal{B}(\mathbb{R})$. Now, each set of the form $(-\infty,t]$ is the union of countably many closed bounded sets, so the $\sigma$ algebra generated by these sets will be contained in $\mathcal{K}(\mathbb{R})$. In other words $\mathcal{B}(\mathbb{R}) \subset \mathcal{K}(\mathbb{R})$, and the result follows.