If you reverse the order of integration (quoting Fubini appropriately) then

.

(The limits of integration change so as to ensure that both repeated integrals are over the triangular region .)

You are told that the integral exists. Also, the function is measurable. It follows that the integral exists.

By Fubini's theorem, exists p.p.(y). But this integral is . So choosing a value of y for which the integral exists, we see that exists. In other words, f is integrable on [0,1].