Proof of uniform continuity of the convolution. Assume throughtout that $1 \leq p &lt; \infty$ and $1&lt; q \leq \infty$ and $\frac{1}{p} + \frac{1}{q} = 1$ and $m$ is the lebesge measure. I have to prove the following theorem:


  Theorem 2: If $f \in L^p(\mathbb{R}^n)$ and $g \in L^q(\mathbb{R}^n)$, then $f \star g$ is uniformly continuous.


We saw the following theorem in the lecture with proof:


  Theorem 1: If $1 \leq p &lt; \infty$ and $1&lt; q \leq \infty$ and $f: \mathbb{R}^n \rightarrow \mathbb{R}$ is locally $p$-integrable, then $$\lim_{\rho \rightarrow 0} \int_K |f(x+\rho) -f(x)|^pdm = 0$$
  for any Lebesgue measurable subset $K \in \mathbb{R}^n$.


My attempt to prove uniform continuity is the following: 

Let $\epsilon &gt; 0$. Since I can use Theorem 1, there exists a $\delta &gt; 0$ such that for all $|\rho| &lt; \delta$ we have $$\int_{\mathbb{R}^n}|f(y+\rho)-f(y)|^pdm &lt; (\frac{\epsilon}{||g||_{L^q}+1})^p$$

Now let $x,\rho \in \mathbb{R}^n$ with $|\rho|&lt;\delta$. The convolution $(f \star g)(x)$ is defined as an integral $\int f(x-y)g(y)dm(y)$ outside of a set $A$, and as $0$ inside $A$, where A is the set where $f(x-y)g(y)$ is not Lebesgue integrable with respect to $y$. If $x+\rho$ and $x$ are in $A^c$, then I can estimate
\begin{align*}
|(f \star g)(x + \rho)-(f \star g)(x)| &amp;= |\int_{\mathbb{R}^n}(f(x + \rho -y)-f(x-y))g(y)|dm(y) \\
&amp;\leq ||f(x+\rho - \cdot)-f(x- \cdot)||_{L^p}||g||_{L^q} \\
&amp;=(\int_{\mathbb{R}^n}|f(y+\rho)-f(y)|^pdm(y))^{1/p}||g||_{L^q} \\
&amp;&lt; \frac{\epsilon}{||g||_{L^q}+1}||g||_{L^q} \\
&amp;&lt;\epsilon.
\end{align*} .
This estimate works now fine if I assume that $x+\rho$ and $x$ are in $A^c$ where I can write the convolution as an Integral. And if $x+\rho$ and $x$ are in $A$, then the convolution is equal to $0$ and thus $$|(f \star g)(x + \rho)-(f \star g)(x)|=0 &lt; \epsilon$$.

But if I have the cases 


$x+\rho \in A$ and $x \in A^c$ 
$x+\rho \in A^c$ and $x \in A$


then I can no longer write the convolution $f \star g$ as an integral. How do I save the proof? Or do I have to use a completely different argument to include those cases?
The proof is actually correct.
Holder inequality gives that: $\ \ \forall x\in R^n,\ \ \int_{R^n}{\left|f(x-y)g(y)\right|\le\left(\int_{R^n}\left|f\right|^p\right)^\frac{1}{p}}\left(\int_{R^n}\left|g\right|^q\right)^\frac{1}{q}&lt;+\infty$
This shows that A is empty, and your proof is correct.
The proof is actually correct.
Holder inequality gives that: $\ \ \forall x\in R^n,\ \ \int_{R^n}{\left|f(x-y)g(y)\right|\le\left(\int_{R^n}\left|f\right|^p\right)^\frac{1}{p}}\left(\int_{R^n}\left|g\right|^q\right)^\frac{1}{q}&lt;+\infty$
This shows that A is empty, and your proof is correct.