Suppose $f \in L^1(\mathbb{R})$ with Lebesgue measure and $r > 0$. Does $f(rx)$ converges to $f$ in $L^1$ as $r \rightarrow 1$ ? Put differently, does $$ \| f(rx) - f(x)\|_1 \rightarrow 0$$ as $r \rightarrow 1$.
Using Lebesgue dominated convergence, we can pass the limit under the integral sign but then I'm stuck because it doesn't feel right to expect $$ |\lim\limits_{r \rightarrow 1} f(rx) - f(x)\big| = 0$$ whenever $f$ is not continuous at $x$.
Elaborating on my comment, I believe the answer is yes. Recall that the continuous, compactly supported functions are dense in $L^1$: that is, for any $f \in L^1(\mathbb{R})$ and $\epsilon>0$, we can find $g$ continuous with compact support such that $\vert\vert g-f \vert\vert_1<\epsilon$. Suppose that you proved the result for continuous functions with compact support.
With that in mind, fix $\epsilon>0$ and choose $g$ such that this holds for our $f$. Now, we can compute, using the Triangle Inequality: \begin{align} \int \vert f(rx)-f(x) \vert&=\int \vert f(rx)-g(rx)+g(rx)-g(x)+g(x)-f(x)\vert\\ &\leq \int \vert f(rx)-g(rx)\vert +\int \vert g(rx)-g(x)\vert +\int \vert g(x)-f(x)\vert\\ &< \frac{1}{r}\int \vert f(u)-g(u)\vert+\int \vert g(rx)-g(x)\vert+\epsilon\\ &<\frac{(1+r)\epsilon}{r}+\int \vert g(rx)-g(x)\vert \end{align} The first inequality is just the Triangle Inequality, and the second follows from assumption and using the substitution $u=rx$ (I believe this holds for the Lebesgue Integral). The final integral can be made arbitrarily small if you have proved it for continuous functions, so the bound can be made arbitrarily small by choosing $r$ small first so that the last integral is as small as you like, and then letting $\epsilon$ go to 0.