Suppose
$f_n, f : \mathbb{R}^d \to \mathbb{C}$ are bounded, uniformly continuous, and go to zero at infinity.
$|f_n| \leq |f|$
$f_n \to f$ pointwise
$0 \leq \int f_n \leq 1$
$\int f_n$ converges
$\int |f_n| \to \int |f|$
Can we conclude $\int f_n \to \int f$?
Note that $\int |f| < \infty$ is not assumed, so it is not obvious if Fatou or dominated convergence can be applied here.
For $n\in\mathbb N$ let $$ f_n(x) = \begin{cases} \frac{\sin(x) }{k}, & (k-1)\pi \le |x| \le k\pi, k=1,\dotsc, n, \\ 0, & \text{otherwise.} \end{cases}$$ It is easily to see that:
$f_n$ has compact support, and admits the Lipschitz constant $1$.
$f_n$ converges uniformly to $f$, defined by $$ f(x) = \frac{\sin(x) }{k} $$ for $(k-1)\pi \le |x| \le k\pi, k\in\mathbb N$. In particular, $f(x)\to 0$ as $|x|\to 0$ and also admits the Lipschitz constant $1$.
$|f_n|$ increases to $|f|$.
$\int f_n = 0$ and thus converges
$\int |f_n| = 4 \sum_{k=1}^n k^{-1} \to \infty = \int |f|$.
As $\int |f| = \infty$, $\int f$ isn't well-defined.
N.B. There is a result stating: If $f_n$ and $f$ are integrable and $f_n\to f$ a.e. Then, $f_n \to f$ in $L^1$ if and only if $\int |f_n| \to \int |f|$.