$f(x)$ is a itegrable function on $(0,+\infty)$ and $\int_{0}^{+\infty}f(x)dx$ coverge. Prove that $$\lim_{t\to 0^+}{\int_{0}^{+\infty }e^{-tx}f(x)dx}=\int_{0}^{+\infty }f(x)dx$$
I got that problem when I learned about improper-integals with a parameter. $\int_{x\to 0^+}^{+\infty}e^{-tx}f(x)dx$ is uniformly converge but $f(x)$ is not continuous (if we had, it would be easier).
BIG Hint
When $t>0$, $$|e^{-tx}f(x)|\leq |f(x)|.$$
Second way (without DCT)
$$|f(x)e^{-tx}-f(x)|=|f(x)||e^{-tx}-1|.$$ Using MVT, $$|e^{-tx}-1|\leq |t|\underset{t\to 0}{\longrightarrow }0,$$ and thus, the convergence is uniform (in $x$). Therefore, you can permute limit and integral.