For a finite Borel measure on $\mathbb{R}$ and $\mu$, $f \in L^1$ compute $\lim_{n \rightarrow \infty} \int_{0}^1 |f(x)|^{1/n} \, d\mu(x)$. The title says it all: given a finite Borel measure $\mu$ on $\mathbb{R}$ and $f \in L^1$, compute $$\lim_{n \rightarrow \infty} \int_{0}^1 |f(x)|^{1/n} \, d\mu(x).$$
I know that I should be looking to invoke the Dominated Convergence Theorem, but I am having trouble finding a dominating function. If we let $g_{n} = |f(x)|^{1/n}$, my first instinct was to say that $|f(x)|$ is a dominating function but $g_{n}$ is not necessarily less than or equal to $|f|$ (for instance the inequality does not hold if $f(x) \in (0,1)$ and $n \geq 2$). 
If I was able to find a dominating function, then we would have $$\lim_{n \rightarrow \infty} \int_{0}^1 |f(x)|^{1/n} \, d\mu(x)=\int_{0}^1\lim_{n\rightarrow \infty} |f(x)|^{1/n} \, d \mu(x),$$
by the Dominated Convergence Theorem. However even if I was able to do this, I am unsure as to handle $\lim_{n\rightarrow \infty} |f(x)|^{1/n}$. 
 Certainly $|f(x)|^{1/n}$ is nonnegative and $\lim_{n \rightarrow \infty} |f(x)|^{1/n} = 1$ when $|f(x)|$ is positive, but I am unsure as to handle the limit if it is the case that $|f(x)| = 0$. Would it be worth while to partition $[0,1]$ into two sets, one in which $|f(x)| > 0$, say $A$, and one in which $|f(x)| = 0$, say $B$? Then we would have 
$$
\int_{0}^1 \lim_{n \rightarrow \infty} |f(x)|^{1/n} \, d \mu(x) = \int_{A} \lim_{n \rightarrow \infty} |f(x)|^{1/n} \, d \mu(x) + \int_{B} \lim_{n \rightarrow \infty} |f(x)|^{1/n} \, d \mu (x) = \int_{A} |f(x)|^{1/n} \, d \mu (x),
$$
because $|f(x)|^{1/n} = 0$ for all $n \geq 1$ on $B$. Then at this point $\lim_{n \rightarrow \infty} |f(x)|^{1/n} = 1$ on $A$, and so the
$$
\int_{0}^1 \lim_{n \rightarrow \infty} |f(x)|^{1/n} \, d \mu(x) = \mu(A). 
$$
So my main areas of concern/questions I have: how could one find a dominating function? Could $|f|$ be tweaked in some way (such as through scalar multiplication of something along those lines) so that it dominates the $g_{n}$? Also, does the second portion of this post (regarding the computation operating the assumption of the existence of a dominating function) seem correct? 

Consider each of $\int_{A(i)} |f(x)|^{1/n}d\mu x$ for $i\in \{1,,2,3\}$ where  $A(1)=f^{-1}\{0\}$ and $A(2)=\{x: 0<|f(x)|\leq 1\}$ and $A(3)=[0,1]\setminus (A\cup B).$ 
We have $\int_{A(1)}|f(x)|^{1/n}d\mu x=0.$ 
The sequence $(|f|^{1/n})_n$ is increasing on $A(2)$ and bounded above by $1$ and bounded below by $0.$ 
The sequence $(|f|^{1/n})_n$ is decreasing on $A(3)$ and bounded below by $1$ and bounded above by $|f|.$
