Convergence in $L^p_\sigma$ where $\sigma$ is Radon measure Let $K$ be a compact set in $\mathbb R^n$.
Suppose we have a Radon measure $\sigma$ and denote the Lebesgue measure by $m$.
Suppose we have $f\in L^p_\sigma(K):=\{g \text{ is measurable}\,\vert \,\int_K \vert g \vert^p \,d\sigma <\infty \}$.
I want something like $\int_K \vert f \vert^p \,dm \le C \int_K \vert f \vert^p \,d\sigma$ for any $f\in L^p_\sigma(K)$, where $C$ is some positive constant. Is this possible to hold? And if it holds, can you give me a hint on the proof of it?
Or alternatively, does this weaker claim holds? Weaker Claim: $\int_K \vert f \vert \,dm \le C \int_K \vert f \vert^p \,d\sigma$ for any $f\in L^p_\sigma(K)$.

My attempt: I have considered the Riesz–Markov–Kakutani representation theorem (which relates Radon measures to Riemann-Stieltjes integrals), but our function $f$ may neither be continuous nor have compact support. So this theorem may not apply. I am quite confused now. Thanks for any help.
 A: Take $\sigma = \delta_0$ and assume that $0∈ K$. Then for any $f∈ C^0$,
$$
\int_K |f|^p \,\delta_0(\mathrm d x) = |f(0)|^p
$$
However, take $f_n(x) = \min(n,|x|^{-1/2p})$. Then
$$
\int_K |f_n|^p \,\delta_0(\mathrm d x) = n^p \underset{n\to\infty}{\longrightarrow} \infty
$$
but
$$
\int_K |f_n|^p \,\mathrm d x \underset{n\to\infty}{\longrightarrow} \int_K |x|^{-1/2} \,\mathrm d x < \infty
\\
\int_K |f_n| \,\mathrm d x \underset{n\to\infty}{\longrightarrow} \int_K |x|^{-1/2p} \,\mathrm d x < \infty
$$
It proves that the inequality you are looking for does not hold in general, even your "weaker" version. The only possibility is when $\sigma$ is absolutely continuous with respect to the Lebesgue measure (or at least its positive part, you can always add a general Radon measure as a negative part). Indeed, if your inequality is true, then for any smooth function compactly test function $\varphi∈ C^\infty_c(K)$,
$$
\langle \sigma,\varphi\rangle \leq C\, \langle 1,\varphi\rangle
$$
which is equivalent to say that $\sigma ≤ C$ in the sense of distributions, and so $\sigma$ is absolutely continuous with respect to the Lebesgue measure, and its associated function $g∈ L^\infty$. Reciprocally, if $σ(\mathrm d x) = g(x)\,\mathrm d x$ for some $g∈ L^\infty$, then
$$
\int_K |f(x)|^p\, \sigma(\mathrm d x) = \int_K |f(x)|^p\, g(x)\,\mathrm d x ≤ \|g\|_{L^\infty} \,\int_K |f(x)|^p\,\mathrm d x
$$

Remark: I used the notation $\sigma(\mathrm d x)$ instead of $\mathrm d \sigma$ to avoid confusions with the Stieltjes integrals notation, as explained in my answer here.
