Sufficient condition for measure convergence on the Borel set of $\mathbb{R}^d$ This question follows a previous one:

Let $(\mu_n)_{n\geqslant 1}$ and $\mu$ be $\sigma$-finite measures on $(\mathbb{R}^d,\mathscr{B}(\mathbb{R}^d))$ such that $\forall n\geqslant 1$, $\mu_n(\mathbb{R}^d)\leqslant 1$ and $\mu(\mathbb{R}^d)\leqslant 1$. Let $\lambda$ denote the Lebesgue measure on $\mathbb{R}^d$. Suppose that
  $$\forall n\geqslant 1\quad\exists g_n\in\mathbb{L}_1(\lambda)\quad\forall A\in\mathscr{B}(\mathbb{R}^d)\quad\mu_n(A)=\int_Ag_n\,\mathrm{d}\lambda\tag{1}$$
  and
  $$\exists g\in\mathbb{L}_1(\lambda)\quad\forall A\in\mathscr{B}(\mathbb{R}^d)\quad\mu(A)=\int_Ag\,\mathrm{d}\lambda.\tag{2}$$
  Is it true that, if $$g_n\xrightarrow[n\to+\infty]{}g\qquad \text{$\lambda$-a.e.}\tag{3}$$ and $$\int_{\mathbb{R}^d}g_n\,\mathrm{d}\mu_n\xrightarrow[n\to+\infty]{}\int_{\mathbb{R}^d}g\,\mathrm{d}\mu\tag{4},$$ then
  $$\sup_{A\in\mathscr{B}(\mathbb{R}^d)}|\mu_n(A)-\mu(A)|\xrightarrow[n\to+\infty]{}0~?\tag{5}$$
  And if it is, how does one establish this result ?

If I replace (4) by $$\int_{\mathbb{R}^d}g_n\,\mathrm{d}\lambda\xrightarrow[n\to+\infty]{}\int_{\mathbb{R}^d}g\,\mathrm{d}\lambda\tag{4'},$$ does the result hold ? And if it does, how to prove it ?
 A: By "measure" I assume the definition includes $\mu_n(A) \ge 0$.  Thus $g_n \ge 0$ a.e.
With assumption (4') it does hold, but not with (4).  Not even with $d=1$.
proof with (4') 

Theorem. Let $\Omega$ be a set, $\mathcal F$ be a sigma-algebra on $\Omega$, $\lambda$ be a measure on $(\Omega, \mathcal F)$.  Let $g : \Omega \to [0,+\infty)$ and $g_n : \Omega \to [0,+\infty)$ for $n=1,2,3,\dots$ be measurable functions.  Assume
  $$
\int_\Omega g\;d\lambda < +\infty,\qquad
\int_\Omega g_n\;d\lambda < +\infty, n =1,2,3,\dots
$$
  and
  $$
g_n \to g\quad\text{$\lambda$-a.e.}
$$
  and
  $$
\int_\Omega g_n\;d\lambda \to \int_\Omega g\;d\lambda.
$$
  Then
  $$
\int_\Omega |g_n-g|\;d\lambda \to 0.
$$

Of course the required conclusion follows, since for every $A \in \mathcal F$,
$$
\left|\int_A g_n\;d\lambda - \int_A g\;d\lambda\right|
\le \int_\Omega |g_n-g|\;d\lambda
$$
The proof is in stages.  The first one is the essential step.  For a set $B \in \mathcal F$, write $B^c := \Omega \setminus B$ for its complement.
Step 1.  Proof in the special case $\lambda(\Omega) = 1$ and $g=1$ a.e.  The proof uses Egorov's Theorem.  Let $\varepsilon > 0$ be given.  There exists $B \in \mathcal F$ and $N \in \mathbb N$
such that $\lambda(B^c) < \varepsilon$ and $|g_n(x) - g(x)|<\varepsilon$ for all $x \in B$ and all $n \ge N$.  Also, $\int_\Omega g_n\;d\lambda \to \int_\Omega g\;d\lambda = 1$, so we may also assume 
$\int_\Omega g_n\;d\lambda < 1+\varepsilon$ for $n \ge N$.
So, for $n \ge N$,
$$
\int_B |g_n-g|\;d\lambda \le \varepsilon\lambda(B) \le \varepsilon,
\\
\int_{B^c} g\,d\lambda = \lambda(B^c) < \varepsilon
\\
\int_{B}g_n\;d\lambda \ge \int_B (1-\varepsilon)\;d\lambda 
\ge (1-\varepsilon)^2
\\
\int_{B^c} g_n\;d\lambda = 
\int_\Omega g_n\;d\lambda - \int_{B}g_n\;d\lambda
\le (1 +\varepsilon) - (1-\varepsilon)^2 = 3\varepsilon-\varepsilon^2 < 3\varepsilon
$$
So that
$$
\int_\Omega |g_n-g|\;d\lambda = \int_B |g_n-g|\;d\lambda
+ \int_{B^c} |g_n-g|\;d\lambda
\le 4\varepsilon + \varepsilon = 5\varepsilon
$$
QED
Step 2.  Proof in the case $g > 0$ a.e.
Write
$$
m := \int_\Omega g\;d\lambda
\\
d\tilde{\lambda} := \frac{g}{m}\;d\lambda
\\
\tilde{g}_n := \frac{g_n}{g}
$$
(We may assume $\lambda$ is not the zero measure, and thus that $m \ne 0$.)
Then apply Step 1 with $1,\tilde{g}_n, \tilde{\lambda}$ in place of 
$g, g_n, \lambda$.
Step 3.  Proof in the case $\Omega$ is $\sigma$-finite.
There is a measurable function $h :\Omega \to [0,\infty)$ with $h>0$ and $\int h\;d\lambda = 1$.  Apply Step 2 with $g+h, g_n+h$ in place of
$g, g_n$.
Step 4.  General case.  The set
$$
\widetilde{\Omega} := \{x : g(x) > 0\} \cup \bigcup_{n=1}^\infty
\{x : g_n(x) > 0\}
$$
is $\sigma$-finite.  Apply Step 3 with $\widetilde{\Omega}$ in place of $\Omega$.
counterexample with (4)
Assumption (1) means that $g_n$ is the Radon-Nikidym derivative
$$
g_n = \frac{d\mu_n}{d\lambda}.
$$
So for a function $f$ we have
$$
\int_A f\;d\mu_n = \int_A f g_n\;d\lambda.
$$
Thus, the rather unnatural assumption (4) is exactly
$$
\int_{\mathbb R} g_n^2\;d\lambda \to \int_{\mathbb R} g^2\;d\lambda.
$$
So the idea is: choose $g_n$ so that $\int g_n^2\;d\lambda \to \int g^2\;d\lambda$ but not $\int g_n\;d\lambda \to \int g\;d\lambda$.
Let $g(x) = 0$ on $\mathbb R$.  So $g(x) \ge 0$ and $\int_{\mathbb R} g\;d\lambda = 0 \le 1$.  
Let 
$g_n(x) = \frac{1}{n}$ on $[n,2n]$ and $0$ elsewhere.  So $\int_{\mathbb R} g_n\;d\lambda = 1 \le 1$. And also $g_n \to 0 = g$ a.e.  But for $A=\mathbb R$, we have $|\int_A g_n\;d\lambda - \int_A g\;d\lambda|=1$, so (5) fails.
Now check (4).  $\int_{\mathbb R} g_n^2\;d\lambda = \frac{1}{n} \to 0 = \int_{\mathbb R} g^2\;d\lambda$.
