# A Riemann Integrability Question

Define $f:\mathbb{R} \rightarrow \mathbb{R}$. For any fixed closed interval $[a,b]$,$f(x)$ is $Riemann$ integrable on $[a,b].$

Show that $\forall a,b;c,d\in\mathbb{R},a<b,c<d.$$\int_{a}^{b}dx\int_{c}^{d}f(x+y)dy=\int_{c}^{d}dy\int_{b}^{a}f(x+y)dx.$

If $f (x+y)$ is $Riemann$ integrable on $\mathcal{R}=[a,b]\times [c,d]$,then we can easily get the equality by applying Fubini's theorem. So the key to this question is to ensure that $f (x+y)$ is Riemann integrable on $[a,b]\times [c,d]$ .Let $A= \lbrace (u,v)\in \mathcal {R^{\circ}} \quad |\quad f (x+y) \text{ is discontinuous at } (u,v)\rbrace$,we only need to prove $A$ has Lebesgue measure zero .But how can I prove $m(A)=0$ is ture?

• Do you mean "proof" or did you mean really mean to say "prove"? – Gregory Grant Jun 11 '15 at 12:38
• Your question doesn't make any sense. $f$ is defined on $\mathbb R$ so what do you mean by "$f(x+y)$ is integrable on $[a,b]\times [c,d]$" (which is a subset of $\mathbb R^2$? – Math1000 Jun 11 '15 at 13:44
• @Math1000: Of course it makes sense: he speaks of the function $(x,y) \mapsto f(x+y)$. – Alex M. Jun 11 '15 at 13:50
• @AlexM. That isn't clear from the question. – Math1000 Jun 11 '15 at 13:53

HINT: Consider the set of discontinuities of $f$ in the interval $[a+c,b+d]$. Cover it by a countable number of intervals whose total length sums to less than $\epsilon$. For each interval $I_j$, consider the set $A_j = \{(u,v): u+v\in I_j\}$. Show that the sum of the areas of the sets $A_j$ is less than some (fixed) multiple of $\epsilon$.
• I am sorry to trouble you.I am failed to prove "the sum of the areas of the sets $A_{j}$ is less than some (fixed) multiple of $\epsilon$." May I have some further hints ? I find another proof of my question applying the definition of Riemann integrability straightforward ,but this proof is not concise. – Elliot Jun 12 '15 at 2:36
• Yes, there are definitely more direct routes. Draw a picture. $A_j$ is a region inside the rectangle of the form $r_1<u+v<r_2$, and this has area at most $(r_2-r_1)\max(d-c,b-a)$. – Ted Shifrin Jun 12 '15 at 2:52