integral is equal to sum of integrals

Here is a proof that uses only Riemann integration theory: since $f$ is continuous and $f(x) <1$ for all $x$ it follows that the maximum value of $f$ is less than $1$. Now you can see that the partials sums of the series $\sum (f(t))^{n}$ converge uniformly. Using the following result you can conclude the proof:

If $f_n$, $f$ are continuous and $f_n \to f$ uniformly then $\int f_n(x) \to \int f(x)dx$

Let $\varepsilon > 0$, $M = \max_{x \in [0,1]}f(x)$, and $I = \int_0^1 \frac{\mathrm{d}t}{1-f(t)}$. Then, for any integer $N > 0$, \begin{align*} &\left| \sum_{n=0}^\infty \int_0^1 f(t)^n \, \mathrm{d}t - \int_0^1 \frac{\mathrm{d}t}{1 - f(t)} \right| \\ &\qquad = \left| \sum_{n=0}^{N-1} \int_0^1 f(t)^n \, \mathrm{d}t + \sum_{n=N}^\infty \int_0^1 f(t)^n \, \mathrm{d}t - \int_0^1 \frac{\mathrm{d}t}{1 - f(t)}\right| \\ &\qquad = \left| \int_0^1 \sum_{n=0}^{N-1} f(t)^n \, \mathrm{d}t + \sum_{n=N}^\infty \int_0^1 f(t)^n \, \mathrm{d}t - \int_0^1 \frac{\mathrm{d}t}{1 - f(t)} \right| \\ &\qquad = \left| \int_0^1 \frac{1 - f(t)^N}{1 - f(t)} \,\mathrm{d}t + \sum_{n=N}^\infty \int_0^1 f(t)^n \, \mathrm{d}t - \int_0^1 \frac{\mathrm{d}t}{1 - f(t)} \right| \\ &\qquad = \left| \int_0^1 \frac{\mathrm{d}t}{1 - f(t)} - \int_0^1 \frac{f(t)^N}{1 - f(t)} \, \mathrm{d}t + \sum_{n=N}^\infty \int_0^1 f(t)^n \, \mathrm{d}t - \int_0^1 \frac{\mathrm{d}t}{1 - f(t)} \right| \\ &\qquad = \left| - \int_0^1 \frac{f(t)^N}{1 - f(t)} \,\mathrm{d}t + \sum_{n=N}^\infty \int_0^1 f(t)^n \, \mathrm{d}t \right| \\ &\qquad \leq \left| \int_0^1 \frac{f(t)^N}{1 - f(t)} \,\mathrm{d}t \right| + \left| \sum_{n=N}^\infty \int_0^1 f(t)^n \, \mathrm{d}t \right| \\ &\qquad \leq \left| M^N I \right| + \left| \sum_{n=N}^\infty M^n \right| \\ &\qquad = M^N I + \frac{M^N}{1-M} \text{,} \end{align*} which is less than $\varepsilon$ as long as we choose $N$ large enough that $$ M^N < \frac{\varepsilon}{I + \frac{1}{1-M}} \text{.} $$ Since $0 \leq M < 1$, such a choice of $N$ is always possible.