Assume: $\mathscr L(\Bbb R^n)$ is the set of Lebesgue-integrable functions, $vol_n(X)=\int_{\Bbb R^n} 1_X$, and $\int^*f=\inf\{\int h: h\in S_*, 1_X\le h\}$.
Let $X\subset\Bbb R^n$ and $\varepsilon>0$ such that $1_X\in\mathscr L(\Bbb R^n)$. Then there exists an open set such that $X\subset\Omega$ and $vol_n(\Omega)<vol_n(X)+\varepsilon$.
I have other problems that are alike, and I'm using some bits of their proofs, but this one is giving me a hard time.
We have that, since $1_X\in\mathscr L(\Bbb R^n)$ then $\int 1_X=\int^*1_X$, then there exists $g\in S_*$, ie, $g$ is lower semicontinuous, and $1_X\le g$.
Now lets consider the open set: $(1-\varepsilon,\infty)$, then $g^{-1}(1-\varepsilon,\infty)$ is open. If $x\in X$ then $g(x)\ge 1_X(x)=1\gt1-\varepsilon$, then $g(x)\gt1-\varepsilon$ then $x\in g^{-1}(1-\varepsilon,\infty):=\Omega$, so $X\subset\Omega$. I got that open set from modifying another problem, so it might be inaccurate.
Now we basically have to prove that $\int^*1_{\Omega\setminus X}<\varepsilon$.
I tried to write $\int^*1_{\Omega\setminus X}$ in terms of $g$, but it didn't helped, and also in terms of $\int 1_X$ and also didn't helped, I also tried to find a bound $\delta(\varepsilon)>0$, such that $\int g<\delta$, but I get a lot of trouble when trying to show somethin like this: $$1_{\Omega}\le g\frac{1}{\delta}\\ \Rightarrow 0\le \int 1_{\Omega}\le \int g\frac{1}{\delta}<\varepsilon$$ what can I do? Intuitively I see that the set $\Omega$ is somehow very close to the set $X$, is $\varepsilon$ close, I think.
Edit: $S_*:=\{f:\Bbb R^n\to\Bbb R\cup\{\infty\}| f \text{ is lower semicontinuous and $\exists K$ compact such that $f(x)\ge0 \forall x\in \Bbb R^n\setminus K$ }\}$
$S^*:=\{f:\Bbb R^n\to\Bbb R\cup\{\infty\}| f \text{ is upper semicontinuous and $\exists K$ compact such that $f(x)\le0 \forall x\in \Bbb R^n\setminus K$ }\}$
$\int^*f=\inf\{\int h: h\in S_*, 1_X\le h\}$
$\int_*f=\inf\{\int h: h\in S^*, 1_X\ge h\}$
Then a function $f:\Bbb R^n\to\Bbb R\cup\{\pm\infty\}$ is lebesgue-integrable iff $$\int_{\Bbb R^n}f:=\int_*f=\int^*f\in\Bbb R$$
I'm not familiar with what precisely you're assuming, but here's an attempt from what it seems you're allowed to work with.
Fix $\epsilon > 0$ and let $g : \mathbb{R} \to \mathbb{R}$ be lower semicontinuous such that $vol(X) \leq \int g \leq vol(X) + \epsilon$. Since $\int 1_X = \int^* 1_X$, it seems that you can always do this by the formula you gave for $\int^* 1_X$.
Define $\Omega = g^{-1}(1-\epsilon, \infty)$, which is open, and so (assuming the monotonicity of the Lebesgue integral and homotheticity, i.e. pulling out constant terms from the integrand), $$ (1 - \epsilon) vol(\Omega) \leq (1 - \epsilon) \int 1_{\Omega} \leq \int_{\Omega} g \leq \int g \leq vol(X) + \epsilon $$ So, setting $\epsilon$ sufficiently small (depending on $vol(X)$) should give what you want.