Hypothesis of dominated convergence theorem The dominated convergence theorem says :

Suppose $(f_n)$ is a sequence of measurable function s.t. $f_n(x)\longrightarrow f(x)$ a.e. as $n\to \infty $. If $|f_n(x)|\leq g(x)$ and $g$ is integrable, then $$\lim_{n\to\infty }\int f_n=\int f.$$

Quest 1) If $|f|$ is integrable, do we directly have $\lim_{n\to\infty }\int f_n=\int f$ or we need to have $|f_n(x)|\leq |f(x)|$ ? 
Quest 2) If no, I have that $g$ is continuous with compact support (and thus integrable). Why the dominated convergence theorem allows me to conclude that $$\lim_{\delta\to 1}\int g(\delta x)dx=\int g(x)dx\ \ ?$$
Don't talks about change the variable $u=\delta x$ to prove it, it's not my question (I know how to prove this). My question is precisely : Why can we conclude using dominated convergence ? Because we don't necessarily have $|g(\delta x)|\leq g(x)$, and thus I'm a little annoyed by this argument. 
 A: For question 2:
If $g$ is continuous with compact support $K$ then $g$ is bounded; say $|g|\leq M$.
Since $K$ is compact, it is closed and bounded, and hence has finite measure.
Let's consider the functions $g(\delta x)$ for $\delta$ close to $1$. These functions are supported of $\frac{1}{\delta}K$, and $|g(\delta x)|\leq M$.
So what are these functions dominated by? Suppose
$$K\subset [-N,N],$$ for some large $N$. Then $g(\delta x)$ is dominated by the function 
$$M\chi_{[-N^2,N^2]}$$which is integrable.
Finally since $g$ is continuous, $g(\delta x) \rightarrow g(x)$ as $\delta \rightarrow 1$.
A: For question 1, to apply the dominated convergence theorem, you definitely need $|f_{n}| \leq |f|$ even if $f$ is integrable (as the first comment points out).
Here's a simple example to show why the integrability of $f$ is not enough. Let $f_{n}(x) = 1$ if $x \in [n,n+1)$, and $0$ otherwise. Here the domain of $f_{n}$ is $\mathbb{R}$. Then $f$ is identically equal to zero and obviously integrable, but
$$
1=\lim_{n} \int f_{n} > \int f = 0.
$$
On the other hand, to show $\lim_{n} \int f_{n} = \int \lim_{n} f_{n}$, you don't have to use the dominated convergence theorem. There're some extensions of the theorem that do not require the existence of a dominating function. There's also a recent paper that gives a necessary and sufficient condition for $\lim_{n} \int f_{n} = \int \lim_{n} f_{n}$.
