Given a Radon-measure $\mu$ on $X$, I want to show that for every positive, $\mu$-integrable function $f\colon X\to\mathbb{R}$ and every $\varepsilon>0$ there exists a compact $K\subseteq X$ for which $$\int_{X\setminus K}f\,\mathrm{d}\mu<\varepsilon.$$
Intuitively this should be true, but I have failed to prove it thus far.
In case $\mu(X)<\infty$ I can assure for any $\delta>0$ the existence of a compact $K$, for which $\mu(X\setminus K)<\delta$, so if I could somehow prove the boundedness of $f$ outside $K$ a.e., I'd be done. As far as I know this cannot be done since I can choose the Lebesgue measure on $(0,1]$ and the function $x\mapsto x^{-1/2}$, which is not bounded whatsoever. This is where I am stuck. I much appreciate any help.
Here is a sketch:
Let's start with the case $\mu(X)<\infty$. Let $\langle K_n\rangle$ be an increasing sequence of compact sets such that $\sup_n\mu(K_n)=\mu(X)$. Let $K=\bigcup_n K_n$ and note that $\mu(X\setminus K)=0$. Now $$\int f~\mathrm~d\mu=\int_{X\setminus K} f~\mathrm d\mu+\int_K f~\mathrm d\mu=\int_K f~\mathrm d\mu=\int 1_K f~\mathrm d\mu=\lim_n\int 1_{K_n}f~\mathrm d\mu=\lim_n \int_{K_n} f~\mathrm d\mu,$$ where the limit argument follows from the monotone convergence theorem.
The assumption that $\mu(X)<\infty$ is not needed though. Let $X_n=\{x\in X\mid f(x)\geq 1/n\}$ and note that $\int f~\mathrm d\mu=\lim_n\int_{X_n} f~\mathrm d\mu$ by the monotone convergence theorem. Moreover, each $X_n$ must have finite $\mu$-measure if $f$ is positive and $\mu$-integrable. One can then replace $X$ with $X_n$ for $n$ large enough in the argument above.