$ \text{First, if } f(x)=c \text{, we obtain from integration by parts that } \int_0^1 fd\phi_n = cg(1)^n. \text{ Next, we observe that } $ $ g(1)<1 \Rightarrow |\int_0^1 fd\phi_n| \leq \lim_{||\pi|| \rightarrow 0} \sum_i |f(x^*)|[g(x_i)^n-g(x_{i-1})^n] \leq Mg(1)^n \rightarrow 0, \text{ since f continuous implies f bounded.} $
$ \text{So assume g attains the value one at some point, and let } \alpha = \min \{ x\in[0,1] | g(x)=1 \}. \text{ Next, given } \epsilon > 0 \text{, let } \ \delta $ $ \text{ be such that } |x-\alpha|<\delta \Rightarrow |f(x)-f(\alpha)| < \epsilon. \text{ We note that} $
$ \int_0^{\alpha-\delta} fd\phi_n \rightarrow 0, \text{ by the second observation, and} \int_{\alpha+\delta}^1 fd\phi_n = \int_{\alpha+\delta}^1 fd(1) = 0. $
$ \text{The first observation gives us that } f(\alpha) - \epsilon < \int_{\alpha-\delta}^{\alpha+\delta} fd\phi_n < f(\alpha) + \epsilon\text{. Since }\epsilon \text{ was arbitrary, } \text{this completes the proof. } \square $
