I am trying to show that the condition of bounded by an integrable function is crucial in the Dominated Convergence Theorem.
Consider a sequence of functions $(f_n)$ on $\mathbb{R}, $ which is equipped with the Lebesgue measure, defined by $$f_n(x) = \frac{1}{\sqrt{2\pi}} e^{- \frac{(x-n)^2}{2}}.$$ So, each $f_n$ is the probability density function of the normal distribution with mean $n$ and variance $1.$ Clearly $$\lim_{n\to\infty} \int_{-\infty}^\infty f_n(x) \, dx = 1 \neq 0 = \int_{-\infty}^\infty \lim_{n\to\infty} f_n(x) \, dx.$$ In this case, the Dominated Converegence Theorem fails because $f_n$ is not bounded by an integrable function. Intuitively, this is clear as $f_n$ is 'running' towards infinity. However, I have difficulty showing it.
Suppose there is an integrable function $g$ such that $e^{-(x-n)^{2}/2} \leq g(x)$ for all $n$ and $x$. Then $g(x) \geq e^{-1/8}$ on the interval $(n-\frac 1 2, n+\frac 1 2)$ for each $n$. Hence $\int g(x) dx \geq \sum_n \int\limits_{n-\frac 1 2}^{n+\frac 1 2} g(x) dx =\infty$.