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In Complex and real analysis by Walter Rudin

It is stated that $$\int_{-\infty}^{+\infty} \left| \int_{-\infty}^{+\infty}f(x-y) g(y) \, dy\right| \, dx \le \int_{-\infty}^{+\infty} dx \int_{-\infty}^{+\infty}|f(x-y) g(y)| \, dy$$

where $f \in L^1(R^1), g \in L^1(R^1) $ and we already know that the function $y \rightarrow f(x-y)g(y)$ and the function $x \rightarrow \int_{-\infty}^{+\infty} f(x-y)g(y) \, dy$ are integrable so we can use Fubini.

But even using Fubini I can't seem to make the product of integrals appear, how is this done?

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    I think this is a typo; the $dx$ on the right side seems to be misplaced. Since $\int_{-\infty}^\infty dx$ does not exist. I guess this inequality is just saying $|\int \cdot | \leq \int |\cdot |$. – 2017-01-03
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    I don't understand the question. You are aware of the fact that the integral on the right hand side is just another way of writing down $$ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |f(x-y)g(y)| dy dx$$? – 2017-01-03
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    @Thomas To be honest I have also never seen this notation. Like writing $\int dx (x^2+1)$. – 2017-01-03
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    @MarvinF. It's not as common as the notation in my comment, but it is used in several textbooks. – 2017-01-03
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    @Thomas Okay thank you for the clarification. Good to know if I ever stumble upon something like this. (Like now) – 2017-01-03
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    @MarvinF. I learned this in a similar way. In some textbooks in physics it seems to be the standard even. – 2017-01-03

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