The harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ diverges, because otherwise the functions $f_n := \frac{1}{n} 1_{[0,n]}$ would be dominated by an absolutely integrable function. But $$\int_{\bf R} \lim_{n \to \infty} f_n(x)\ dx = 0 \neq 1 = \lim_{n \to \infty} \int_{\bf R} f_n(x)\ dx$$, dx,$$contradicting the dominated convergence theorem. 2 added 2 characters in body An example that came up in my measure theory class today: The harmonic series \sum_{n=1}^\infty \frac{1}{n} diverges, because otherwise the functions f_n := \frac{1}{n} 1_{[0,n]} would be dominated by an absolutely integrable function. But \int_{\bf \int_{\bf R} \lim_{n \to \infty} f_n(x)\ dx = 0 \neq 1 = \lim_{n \to \infty} \int_{\bf R} f_n(x)\ dxdx$$, contradicting the dominated convergence theorem.
The harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ diverges, because otherwise the functions $f_n := \frac{1}{n} 1_{[0,n]}$ would be dominated by an absolutely integrable function. But $\int_{\bf R} \lim_{n \to \infty} f_n(x)\ dx = 0 \neq 1 = \lim_{n \to \infty} \int_{\bf R} f_n(x)\ dx$, contradicting the dominated convergence theorem.