Limit of $\int_{-\infty}^{\infty} f(x)\sin(tx)dx$ as $|t|\to\infty$

Let $f$, a Lebesgue integrable function in $\mathbb{R}$ ($\int_{\mathbb{R}}|f| < \infty$). Let: $$g(t) := \int_{-\infty}^{\infty} f(x)\sin(tx)dx$$ Show that $g$ is continuous (which I did), and that: $$\lim_{|t| \rightarrow \infty} g(t) = 0$$

Why is the second part correct?

Thanks!

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this may help. AD.'s answer, in particular. –  David Mitra Feb 25 '12 at 17:55
This is so called Riemann-Lebesgue Lemma! –  беркай Mar 20 '12 at 20:58

The result is trivial if $f$ is the characteristic function of an interval, and therefore also if it is a step function. Can you take it the rest of the way from there?

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I can see why it is true for step-functions, and we can use the fact that the set of step-functions is dense in a closed interval, but how do we know whether or not there is a sequence of step-functions converging pointwise to f over the real line? Can we apply DCT? –  The Substitute Feb 9 at 7:00
@TheSubstitute I think using the DCT may be a bit tricky. But you can approximate any $L^1$ function with a simple function in the $L^1$ norm, and likewise any simple function with a step function (because any measurable set of finite measure can be approximated with a finite union of intervals). There are quite a number of details to fill in, and the step from approximating in the $L^1$ norm to proving the theorem for $L^1$ functions is a common stumbling block for beginning analysis students. –  Harald Hanche-Olsen Feb 9 at 14:53