By the Extreme Value Theorem, continuous on with is bounded.

The proof proceeds by contrapositive: if is not integrable, then is not continuous. If is not integrable, there is such that for every partition of ,

hence some has . By the Extreme Value Theorem there are with .

This applies in particular to the uniform partitions : for each pick with and . The sequences and are bounded, so by the Bolzano-Weierstrass Theorem they have convergent subsequences . Their limits satisfy , since

yet along the same indices, so and do not converge to the same limit. Thus is not sequentially continuous, hence not continuous.