# integrable function is also integrable in its sub partitions

Assume that f is integrable on [a,b]. How can we show that f is integrable on every interval $[c,d]\subseteq [a,b]$?

Here is what I've done so far:

If f is integrable on [a,b] then we know for any $\epsilon > 0 \text{ } \exists \text{ a partition on [a,b] such that } U(f,p) - L(f,p) < \epsilon$

If we consider P = [c,d] as a $\textbf{sub-partition of}$ Q = [a,b], then we can say:

$$L(f,P) \le L(F, Q) \le U(F,Q) \le U(F,P)$$

But this actually is useless and does not help us.

From your question it seems that you mean Riemann integrable. One way, which is certainly an overkill, is the use the Riemann-Lebesgue theorem: $f:I\to \mathbb R$, where $I$ is an interval in $\mathbb R$, is Riemann integrable if, and only if, $f$ is bounded and its set of discontinuities has Lebesgue measure $0$. If $J \subseteq I$ is a subinterval of $I$, then the set of discontinuities of $f$ in $J$ is clearly a subset of the set of discontinuities of $f$ in $I$, and since it is it trivial that a subset of a set of Lebesgue measure $0$ has itself measure $0$, the result follows immediately.