Dominated Convergence Theorem assumptions I'm trying to find an example to show that it is impossible to weaken the hypothesis of the dominated convergence theorem that $|f_n|\leq g$ for all $n \ge 1$ with g satisfying   $\int_{X} gd\mu \lt \infty$,
$X$
even if we assume that $μ(X) \lt \infty$.
I'm not actually sure what the question is specifically asking me to do...
 A: Let $X = (0,1)$ and for each $n$ let $f_n(x) = 1/x^2$ if $x \in (1/(n+1),1/n)$ and $0$ otherwise. Then $f_n(x) \to 0$ everywhere on $(0,1)$ but their integrals are bounded away from $0$.
A: You could also look at $f_n(x) = n^3x^n(1-x)$ on $[0,1].$ Here $f_n \to 0$ pointwise on $[0,1]$ while $\int_0^1 f_n \to \infty.$
A: To add to Umberto's answer, the question is basically asking what happens if you weaken the hypothesis for the Dominating Convergence Theorem. In particular, the condition that the dominator function has to have a finite integral is especially important. 
For example, supposed you have a sequence of functions $f_n: [a,b] \rightarrow [0, \infty)$ that takes the integral of $\frac{1}{x}$:

$f_n(x) = \int^{n+1}_1 \frac{1}{x}$

Clearly, there is a dominating function, $g(x) = \int_1^{\infty} \frac{1}{x}$ (which also happens to be the limit function $f = lim f_n$), which is just the sum of the harmonic series, but the measure of its undergraph is not finite. But $\int f_n = \sum_{i = 2}^{n+1} {\frac{1}{i}}$, which is finite, so $\int f_n \not \rightarrow \int f$, so the dominating convergence theorem fails.
To take another example in which neither condition (existence of a dominator, let alone an integrable one) is fulfilled, imagine a sequence of functions $f_n: [a,b] \rightarrow [0,\infty)$ that graph to triangles whose base gets smaller and smaller but the height increases in length. Then the overall planar measure of the undergraph is finite for each $n$, $f_n \rightarrow 0$ pointwise, but the measure of the limit function $f = lim f_n$ is in fact infinite, so $\int f_n \not \rightarrow \int f$. If you want an explicit formula:

$
f_n(x)= 
\begin{cases} 
      n^2x & \text{if  } 0 \leq x \leq \frac{1}{n} \\     
      2n - n^2x & \text{if } \frac{1}{n} \leq x \leq \frac{2}{n} \\          
      0 & \text{o.w.} 
\end{cases}
$

Essentially, the function is constructed so that the graph resembles a triangle of base length $\frac{2}{n}$ and height $n$. If course, the triangle has finite measure for each $n$, but its measure approaches $\infty$ precisely because it does not fulfill the hypothesis at all and thus the conclusion does not hold. But I don't think the question is asking for an example of a sequence of functions with no dominator at all, but I always thought this was a neat example that the full converse (or even the partial converse, based on the example above) of the Dominating Convergence Theorem is not true. 
A: It's actually possible to weaken the hypothesis of the dominated convergence theorem.
For example, let $X=(0,1]$ and $\mu$ be the Lebesgue measure. Let
$$
f_{n}(x) = 
\begin{cases}
[1/n - 1/(n+1)]/n & \text{if $x \in (1/(n+1),1/n]$,}
\\ 0 & \text{otherwise.}
\end{cases}
$$
Then $\int f_{n} = 1/n$ and $\lim_{n} f_{n} = 0$. In this case, there's no dominating function that is integrable, but
$$
\lim_{n} \int f_{n} = \int \lim_{n} f_{n} = 0.
$$
There're various extensions of the theorem, and there's a recent paper that gives a necessary and sufficient condition for the first equality above to be valid. The exact condition is somewhat complicated, but it's strictly weaker than uniform integrability, which is also somewhat complicated and in turn weaker than the hypothesis of the dominated convergence theorem.
