Detail on Portmanteau theorem Let $(\mu_n)_n$ and $\mu$ be probability measures on $(\mathbb{R}^d,B(\mathbb{R}^d))$.
In Portmanteau theorem, one can prove that $(\mu_n)_n$ converges weakly to $\mu$ if and only if for all bounded, lower semicontinuous functions $f$ we have $$\int_{\mathbb{R}^d}f(x)d\mu(x) \leq \liminf_n\int_{\mathbb{R}^d}f(x)d\mu_n(x),$$ this is true because, without loss of generality, we can assume $f \geq 0$ (we can take, for the general case, $f=f^+-f^-$) and using Fatou's lemma, $\liminf_n \mu_n(O) \geq \mu(O)$ for every open $O$ and that $$\int_{\mathbb{R}^d}f(x)d\mu(x)=\int_{0}^{+\infty}\mu(f>x)dx$$ we will obtain the result. 
So, my question is, do we have the same result if we only suppose that $f$ is bounded and lower semi continuous $\mu$ a.e which means that $\mu(Q)=1$ where $Q$ is the set of points where $f$ is lower semicontinuous? 
 A: Yes, this is true. Assume without loss of generality that $f\ge0$, since by the boundedness of $f$, we can always subtract the constant $\inf f$. Define
$$ g_k(x) := \inf\Big\{f(y) + k\|x-y\|\,:\,y\in\mathbb R^d\Big\}.$$
Clearly, $0\le g_k(x) \le g_{k+1}(x)$, and by taking $y=x$ we see $g_k(x) \le f(x)$. Moreover, $g_k$ is continuous: if $x,z\in\mathbb R^d$, then for all $y\in\mathbb R^d$,
$$g_k(x) \le f(y) + k\|x-y\| \le \Big[f(y) + k\|z-y\|\Big] + k\|x-z\|.$$
Taking infimum over $y$, we find
$$ g_k(x) \le g_k(z) + k\|x-z\|,$$
which then implies $|g_k(x)-g_k(z)| \le k|x-z|$, and so $g_k$ is continuous as claimed. Let $g(x)=\lim_{k\to\infty}g_k(x)$. If $x\in Q$ (the set of points where $f$ is lower semicontinuous) and $\epsilon>0$, then there exists $\delta>0$ such that $f(x) \le f(y) + \epsilon$ whenever $\|x-y\|<\delta$. Hence, if $\|x-y\|<\delta$,
$$ f(y) + k\|x-y\| \ge f(y) \ge f(x) - \epsilon,$$
and if $\|x-y\|>\delta$,
$$ f(y) + k\|x-y\| \ge f(y) + k\delta \ge k\delta \ge f(x) - \epsilon $$
for $k$ sufficiently large. This shows, for $k$ sufficiently large, $f_k(x) \ge f(x) - \epsilon$, and so letting $k\to\infty$ we see that $g(x) = f(x)$ for all $x\in Q$.
Now for any $k$, since $g_k$ is continuous, we have
$$ \int g_k\, d\mu = \lim_{n\to\infty} \int g_k\, d\mu_n \le \liminf_{n\to\infty} f \, d\mu_n.$$
Letting $k\to\infty$ and applying the monotone convergence theorem we deduce
$$ \int_{\mathbb R^d} f\, d\mu = \int_Q f\, d\mu = \int_Q g \, d\mu = \lim_{k\to\infty}\int_Q g_k\, d\mu \le \liminf_{n\to\infty} \int_{\mathbb R^d} f\, d\mu_n.$$
This completes the proof.
