Riemann Lebesgue type limit

Question

If $f\in \mathscr{R}[-2,2]$ then show that $$\lim_{h\to 0}\int_{-1}^{1}|f(x+h)-f(x)|\,\mathrm{d}x=0$$ Now, my idea was to approximate by step functions. Given any $\varepsilon >0$ We can find a step function $\varphi$ such that $$\int_{-1}^{1}|f(x)-\varphi(x)|\,\mathrm{d}x<\varepsilon$$ Now, we only need to prove $$\lim_{h\to 0}\int_{-1}^{1}|\varphi(x+h)-\varphi(x)|\,\mathrm{d}x=0$$ for step functions $\varphi$, which I think is obvious, since if $$\displaystyle \varphi=\sum_{i=1}^{n}a_i\mathbf{1}_{[x_{i-1},x_i]}$$ where $x_0=-1, x_n=1$, we can break it into sum of integrals, and prove this for functions constant in an interval, and then the integral is actually $0$. Now, we have, $$\int_{-1}^{1}|f(x+h)-f(x)|\,\mathrm{d}x\le \int_{-1}^{1}|f(x+h)-\varphi(x+h)|\,\mathrm{d}x+\int_{-1}^{1}|\varphi(x+h)-\varphi(x)|\,\mathrm{d}x+\int_{-1}^{1}|\varphi(x)-f(x)|\,\mathrm{d}x$$ but each of the integrals can be made less than $\varepsilon$, which does the proof. But, I am not quite sure about this proof. I think there is some mistake in it. Can someone help me out? Thanks.