$\int_\mathbb{R} \bigg( \frac{1}{h} \int_x^{x+h} |f(t)| dt\bigg) dx= ||f||_{L^1}$? $$\int_\mathbb{R} \bigg( \frac{1}{h} \int_x^{x+h} |f(t)| dt\bigg)  dx= ||f||_{L^1} \;\;?$$ 
I worked out that the equality holds for each $\chi_{[a,b]}$, therefore it holds for each piecewise constant function. By a density argument, it must hold for all functions in $L^1(\mathbb{R})$.


*

*Is this correct?

*If it were true, I feel like there must be an easier argument for this inequality; I just can't quite see it. Maybe using Hardy-Littlewood maximal inequality or Markov's inequality, etc.


Thank you very much for the help!

Your argument is correct, if you note that the left hand side of the equality is a continuous function of $f\in L^1$. Then the denseness perpetuates the equality from the simple functions to all of $L^1$.
But there is an easier argument, change the order of integration:
$$\begin{align}
\int_\mathbb{R} \left(\int_x^{x+h} \lvert f(t)\rvert\,dt\right)\,dx
&= \iint\limits_{x \leqslant t \leqslant x+h} \lvert f(t)\rvert\,dt\,dx\\
&= \iint_{t-h \leqslant x \leqslant t} \lvert f(t)\rvert\,dx\,dt\\
&= \int_\mathbb{R} h\lvert f(t)\rvert\,dt\\
&= h\lVert f\rVert_{L^1}.
\end{align}$$
Since the integrand is non-negative and measurable, the change of order of integration is unproblematic.
