# If a function is integrable on an interval, then it is also integrable on each of its subintervals $[x,y]$

I have to prove the above statement and I tried with this: Since f is integrable function on some interval, let a partition P of that interval. We know that $U(f,P)- L(f,P) < \varepsilon$ by integrability of f. Then let $P_1 = P \cup \{x,y\}$ and also let $P_2 = P \cap [x,y]$. So we have $P_2 \subset P_1$, since it partitions $[x,y]$, and $P_1$ is a refinement of P. Therefore $$U(f,P_2)- L(f,P_2) \leq U(f,P_1)- L(f,P_1) \leq U(f,P)- L(f,P) < \varepsilon$$ And this proves f is integrable on $[x,y]$.

Is this correct and enough? thanks!

You are going in the right direction, but there are some gaps. You want to show that for given $\epsilon>0$, there exists $\delta>0$ such that every partition $P$ of $[x,y]$ that is finer than $\delta$ gives us $U(f,P)-L(f,P)<\epsilon$. And we are given that the corresponding property for the larger interval $[a,b]$ holds.
The idea is indeed that we can use the same $\delta$ for $[x,y]$ that we know exists for $[a,b]$. But: We then need to start from an arbitrary partition of $[x,y]$ that is finer than $\delta$; then extend this to a partition of $[a,b]$ that is still finer than $\delta$; then we can use the inequality chain as you do. (You instead start form a partition of $[a,b]$ and produce a partition of $[x,y]$ from it - but it is not immediately clear that we can reach every (fine) partitoin of $[x,y]$ this way)