Let $f_\epsilon$ converge pointwise to $f$ for $\epsilon \rightarrow 0$ and let $f_\epsilon$ and $f$ be integrable.
Can I swap integral and limes in this case? $$ \lim_{\epsilon \rightarrow 0} \int_\Omega f_{\epsilon}(x) dx = \int_\Omega \lim_{\epsilon \rightarrow 0} f_{\epsilon}(x) dx = \int_\Omega f(x) dx $$
I tried to apply dominated convergence, by using $f + 1$ as the upper bound function, but because of pointwise convergence it does not work.
In general one cannot interchange the limit and integral. By way of a counter example, consider the sequence $(f_n) \subset L^1(\mathbb{R})$ given by $$ f_n(x) := \mathbf{1}_{[n,n+1)}(x). $$ Clearly, $f_n(x) \to 0$ for almost every $x \in \mathbb{R}$. However, \begin{align*} \lim_{n \to \infty} \int_{\mathbb{R}} f_n\,\mathrm{d}m = \lim_{n \to \infty} 1 = 1 \neq \int_{\mathbb{R}} 0\,\mathrm{d}m. \end{align*} Without additional assumptions (e.g. monotonicity or uniform domination by an integrable function), the above shows that one cannot freely pass the limit into the integral.
Even if the measure space is finite one cannot necessarily take the limit inside the integral. Indeed, consider the Lebesgue measure $m$ on $[0,1]$ and define a sequence $g_n$ on $[0,1]$ via the rule $$ g_n(x) := n \mathbf{1}_{\left[0, \frac{1}{n}\right)}(x), \quad \forall n \geq 1. $$ Again, it is clear that $g_n(x) \to 0$ for almost every $x \in [0,1]$. However, as above, $$ \lim_{n \to \infty} \int_{[0,1]} g_n\,\mathrm{d}m = 1 \neq \int_{[0,1]} 0\,\mathrm{d}m = 0. $$
Additional observations
Even when one cannot dominate by an integrable function, there are still situations in which one can pass the limit into the integral. In my experience, the following theorem has turned out to be particularly useful: