Today I was reading Brezis, Functional analysis and there is an exercise about Young inequality. Anyway I would like to have some clarification. Suppose that $1 < p < \infty$ and $1 < q < \infty$, $f \in L^p(\mathbb{R}^N)$ and $g \in L^q(\mathbb{R}^N)$. Let $1/q + 1/p > 1$. Let $r$ be such that $1/p + 1/q - 1 = 1/r$.
In the text it is asked to prove that $ y \to f(x-y)g(y)$ is integrable for almost every $x \in \mathbb{R}^N$, that $f \star g \in L^r$ and $ \Vert f \star g\Vert_r \le \Vert f \Vert_p \Vert g \Vert_q$.
Now, the question is why we can get only almost everywhere integrability with respect to $x$?
I think one could proceed in this way:
One can certainly say that, for each fixed $x$, the function $f(x-y)g(y)$ is measurable. We have, thanks to traslation invariance of lebesgue measure $(\int_{R^N} |f(y)|^p)^{1/p} = (\int_{R^N} |f(x - y)|^p)^{1/p}$. Thus, fixed $x$, $y \to f(x-y)$ is in $L^p$.
Observe that:
$|f(x-y)g(y)| = |f(x-y)|^{p/q'}|g(y)|^{q/p'}(|f(x-y)|^{1-p/q'}|g(y)|^{1-q/p'})$
with:
$|f(x-y)|^{p/q'} \in L^{q'}$
$|g(y)|^{q/p'} \in L^{p'}$
$(|f(x-y)|^{1-p/q'}|g(y)|^{1-q/p'}) \in L^r$
Observe also that $1/{p'} + 1/{q'} \le 1$. We can thus apply the generalized Holder inequality to obtain that $|f(x-y)|^{p/q'}|g(y)|^{q/p'} \in L^{r'}$ ( Observing that $1/{p'} + 1/{q'} = 1 - 1/r = 1/{r'}$).
Now we can apply Holder inequality once more to conclude that $f(x-y)g(y)$ is integrable.
But arguing this way I get integrability for every $x \in \mathbb{R}^N$. Is the reasoning correct?