Let $f_n,\,n\in\mathbb{N}$ be a sequence of real integrable functions, $f_n\to f$ pointwise as $n\to\infty$.
The dominated convergence theorem states that if there exists $g\in L^1$ such that $|f_n(x)|\leq g(x)$ for all $n,x$, then $$ \int f_n(x)\,dx \to \int f(x)\,dx \quad\text{as }\,n\to\infty \;.$$
But now if I have a weaker condition, that is there exists $g\in L^1$ such that $$\int |f_n(x)|\,dx \leq\int g(x) \,dx \;,$$ for all $n$, can I conclude the same?
On
The answer to my question is clearly negative, as Ambram Lipman showed. Anyway the answer become positive if one improves the hypothesis. Precisely assume:
$\bullet$ the integral is done over a bounded domain (or in general against a finite measure $\mu$);
$\bullet$ the sequence $(f_n)_n$ is uniformly bounded not only in $L^1(\mu)$ (as I was asking in my question), but it is uniformly bounded in $L^p(\mu)$ for a $p>1$, that is there exists $C<\infty$ such that for all $n$ $$\int |f_n(x)|^p\,d\mu(x)\leq C \;;$$
then the sequence $(f_n)_n$ is uniformly integrable against the finite measure $\mu$ and therefore $$\int f_n(x)\,d\mu(x) \to \int f(x)\,d\mu(x) \quad\text{as }\,n\to\infty \;.$$
No, you definitely can't. In your hypothesis, you're basically just saying that $\int |f_n(x)| dx$ is uniformly bounded by some constant. All that $g$ does is say that that constant happens to be the integral of some $L^1$ function, but that's true of all constants anyway.
As a counterexample, take $f_n(x)$ to be the indicator function on $[n,n+1]$ (so $f(x) \equiv 0$, but $\int f_n(x) = 1$ for all $n$) and $g(x)$ to be any $L^1$ function with $\int g(x) dx \ge 1$.