Absolute value of an improper integral

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Context: I have been trying to demonstrate that the magnitude of the characteristic function of a random variable $X$, with PDF $f(x)$, is upper bounded by 1. In that process I got this expression:

$$\left \lvert \int_{-\infty}^{\infty} e^{-j\omega x}f(x) dx\right\rvert$$

I have found a proof that shows that

$$\left \lvert \int_{a}^{b} f(x)dx\right\rvert \leq \int_{a}^{b} \lvert f(x) \rvert dx$$

but it is derived considering a closed interval $[a,b]$.

Question: I have tried to find a similar proof when the integral is improper without success. Does that inequality holds for improper integrals? Under what conditions? Do you know a proof of this?

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To prove the inequality for the interval $[a,\infty)$, we may argue as follows.

For a fixed $b > a$, we have $$\left|\int_{a}^{b} f(x) dx\right| \leq \int_{a}^{b}|f(x)| dx \leq \lim_{b \to \infty} \int_{a}^{b} |f(x)| dx = \int_{a}^{\infty} |f(x)|dx$$ where the first inequality holds because of the result you cited; the second inequality is true because the integrand is nonnegative; and the third equality holds by definition of the improper integral.

To summarize, the inequality $$\left|\int_{a}^{b} f(x) dx\right| \leq \int_{a}^{\infty} |f(x)|dx$$ is true for any fixed $b > a$. Therefore, it is true if we take the limit as $b \to \infty$: $$\lim_{b \to \infty}\left|\int_{a}^{b}f(x) dx\right| \leq \int_{a}^{\infty} |f(x)|dx$$ Finally, we can rewrite the left hand side of this inequality as follows: $$\lim_{b \to \infty}\left|\int_{a}^{b}f(x) dx\right| = \left|\lim_{b \to \infty} \int_{a}^{b}f(x) dx\right| = \left|\int_{a}^{\infty}f(x) dx\right|$$ where the first equality is true because the absolute value function is continuous, and the second equality is true by definition of the improper integral.

From the above, we can conclude that $$\left|\int_{a}^{\infty} f(x) dx\right| \leq \int_{a}^{\infty}|f(x)| dx$$

You can argue exactly the same way as $a \to -\infty$ to conclude that $$\left|\int_{-\infty}^{\infty} f(x) dx\right| \leq \int_{-\infty}^{\infty}|f(x)| dx$$ as desired.