Lebesgue Measurable Functions

Question

Let $(X,\mathcal{A},\mu)$ be a measure space with $\mu(X)=1$.$\;$ Suppose $\varphi:X \longrightarrow [0,1)$ is measurable.

Prove that $$\lim_{a\to\infty}\int_{X}\varphi^{a}\:\mathrm{d}\mu=0.$$ My attempt: Define $C_n=\{x \in X \;|\;\varphi(x) < 1- \frac{1}{n} \}.$ From this point i'm stuck.
Any help would be most appreciated.

Answer 1

The simplest approach is to apply the dominated convergence theorem, as hinted to in my comment above. However, if you want to utilize your sets $C_n$, here is a suggestion on how to do so.

First pick $\epsilon > 0$ and observe that the $C_n$ increase to $X$: $$C_1 \subseteq C_2 \subseteq \cdots \quad \text{and} \quad \bigcup_{n=1}^\infty C_n = X.$$ By the upwards continuity of the measure $\mu$, we find $$\mu(X) = \lim_n \mu(C_n)$$ and thus an $N \ge 1$ with the property that $$\mu(X) - \frac{\epsilon}2 < \mu(C_N).$$ In other words, $\mu(X \setminus C_N) < \epsilon/2$. Since $0 \le \varphi^m < 1$ for all $m$, we get the estimate $$\begin{align*} \int_X \varphi^m \, d\mu &= \int_{C_N} \varphi^m \, d\mu + \int_{X \setminus C_N} \varphi^m \, d\mu \\ &\le \int_{C_N} \varphi^m \, d\mu + \mu(X \setminus C_N) \\ &\le \int_{C_N} \varphi^m \, d\mu + \frac{\epsilon}2, \end{align*}$$ and it only remains to show that $\int_{C_N} \varphi^m \, d\mu$ is small when $m$ is large. If $\mu(C_N) = 0$, this is certainly the case, so assume without loss of generality that $\mu(C_N) > 0$. Now, for $x \in C_N$ one has $$0 \le \varphi(x)^m \le \Bigl( 1 - \frac1N \Bigr)^m,$$ so if we pick $M \ge 1$ so large that $$\Bigl( 1 - \frac1N \Bigr)^m \le \frac{\epsilon}{2\mu(C_N)} \quad \text{whenever} \quad m \ge M,$$ we see that $$\int_{C_N} \varphi^m \, d\mu \le \int_{C_N} \frac{\epsilon}{2\mu(C_N)} \, d\mu = \frac{\epsilon}2$$ for such $m$, and we are done.