If $f\in R[a,b]$ then $\lim_{h\to0}\int_{c}^{d}|f(x+h)-f(x)|dx=0$ 
If $f\in R[a,b]$, then prove that $\lim_{h\to0}\int_{c}^{d}|f(x+h)-f(x)|dx=0$, where $a<c<d<b$. $R[a,b]$ is the set of all Riemann Integrable functions on $[a,b]$.

I think the well-known step function corresponding to the lower Riemann sum may work but I am having some doubts. Indeed, given any $\varepsilon>0$ because of integrability of $f$ I can find a partition $P$ of $[a,b]$ such that $\int_{c}^{d}f(x)dx-L(Q,f)<\varepsilon$ where $Q=(P\cup\{c,d\})∩[c,d]$. So I can take $S$ to be my step function corresponding to $L(Q,f)$. The step function $S(x)$ is as follows:
$$S(x)=\inf\{f(x):x\in[x_{i-1},x_i]\},x\in[x_{i-1},x_i]$$
where $Q=\{x_0<x_1<...<x_n\}$.
Then I apply triangle inequality to have,
$$\int_{c}^{d}|f(x+h)-f(x)|dx\leq\int_{c}^{d}|f(x+h)-S(x+h)|dx+\int_{c}^{d}|S(x+h)-S(x)|dx+\int_{c}^{d}|f(x)-S(x)|dx$$
Now the last quantity is less than $\varepsilon$ (by construction). If I can show that there exists $\delta>0$ such that $|h|<\delta$ implies the first two quantities are also less than $\varepsilon$ then I will be done. I believe it happens but am not quite sure as to whether $S(x+h)$ also approximates $f(x+h)$ and whether the second quantity is truly less than $\varepsilon$ (because there may be large jumps for the infima even when the discontinuities fall in a set of measure zero).
I would really appreciate some hints. Let me see if I can pick up from those.
 A: Okay so I have thought on it and have dared to finally post what I feel is a correct solution. But of course, I would expect the esteemed users will verify it.

$f$ being Riemann Integrable on $[a,b]$, we must have, given any $\varepsilon>0$, a continuous function $g$ defined on $[a,b]$ such that $$\int_a^b|f(x)-g(x)|dx<\varepsilon$$ I can write the following by Triangle Inequality $$\int_c^d|f(x+h)-f(x)|dx=$$$$\int_c^d|f(x+h)-g(x+h)+g(x+h)-g(x)+g(x)-f(x)|dx\leq$$$$\int_c^d|f(x+h)-g(x+h)|dx+\int_c^d|g(x+h)-g(x)|dx+\int_c^d|f(x)-g(x)|dx$$
  Now by using the fact that $a<c<d<b$,$$\int_c^d|f(x)-g(x)|dx<\int_a^b|f(x)-g(x)|dx<\varepsilon$$
Clearly, $|g(x+h)-g(x)|<\varepsilon$ for $|h|<\delta$ where $\delta$ is obtained from $\varepsilon$ by continuity of $g$. Hence, $$\int_c^d|g(x+h)-g(x)|dx<(d-c)\varepsilon$$.
Now the first part.$$\int_c^d|f(x+h)-g(x+h)|dx=\int_{c+h}^{d+h}|f(x)-g(x)|dx\leq\int_a^b|f(x)-g(x)|dx<\varepsilon$$

EDIT: I thought that I need to clarify why I basically interchanged the limit and integral in $\int_c^d|g(x+h)-g(x)|dx$. The reason is Uniform Continuity and Uniform Convergence of $g$. The proof is as follows:

$g$ is a bounded, uniformly continuous function, being continuous on $[a,b]$. 
Given any $\varepsilon>0$ there exists $\delta>0$ (which depends only on $\varepsilon$ and not on $x$ by uniform continuity of $g$) such that$$|h|<\delta\implies|g(x+h)-g(x)|<\varepsilon$$ This is true for all $x\in[c,d]$. So,$g(x+h)$ converges uniformly to $g(x)$, as the above is precisely the statement of uniform convergence.
Due to this, we can interchange the limit and integral.

A: Warning: non answer because it likely uses material you are not comfortable with.
You may know that $f \in R[a,b]$ if and only if $f$ is bounded and continuous almost everywhere. In this case the Riemann integral and the Lebesgue integral agree and we are in a position to use dominated convergence to say
$$
\lim_{h \to 0} \int_c^d |f(x+h)-f(x)|dx = \int_c^d \lim_{h \to 0} |f(x+h)-f(x)|dx = \int_a^b 0 dx = 0.
$$
