Is $\int f$ on [0,1] always equal to $\int f$ on [x,1] when take the limit of x to 0? I know that if f is nonnegative, then I can use LMCT to prove it. However, how about f is only bounded, or only the later integral exists and finite? I suspect that if this always holds.
2026-05-03 15:20:09.1777821609
Is $\int f$ on [0,1] always equal to $\int f$ on [x,1] when take the limit of x to 0
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It reduces to showing that $\lim_{x \to 0^+} \int_0^x f(y) dy = 0$. This is trivial in the (proper) Riemann framework, where all functions are assumed bounded. In the Lebesgue framework it requires an approximation step, e.g. the bounded convergence theorem or the dominated convergence theorem.