Arguing that $\int_0^1 f = \lim_{\epsilon \to 0} \int_\epsilon^1 f$

Question

Let $f\ge 0$, my professor said that you could use $\int_0^1 f = \lim_{\epsilon \to 0} \int_\epsilon^1 f$, due to the Monotone Convergence Theorem. In this case, is it because the functions are actually $g_n(x) = \int_{h_n}^1f$ (where $h_n$ is a sequence that tends to $0$)? If so, the MCT states that, if there exist functions $f_1 \leq f_2 \leq ...$ such that $f_n \to f$ then $\int f_n \to \int f$, but in the example with $g_n \to g$ there is no integral (since it is "included" in the function $g_n$).

Answer 1

Using indicator functions, we may write $$\int_\epsilon^1 f = \int_0^1 f \cdot \mathbf{1}_{[\epsilon,1]}$$ Then we may define $g_n(x) = f \cdot \mathbf{1}_{[h_n,1]}$ for some decreasing sequence $h_n$ tending to $0.$