If $a>1$ and $f\in L^a([0,1])$, show that $\|f\|_a\to \|f\|_1$ If $a>1$ and $f\in L^a([0,1])$, show that $\lim_{a\to 1}\|f\|_a=\|f\|_1$.
Proof: Assume $f$ is bounded. $f^a\to f$ pointwise. By the dominated convergence theorem, $\int|f|^a\to\int|f|$.
I'm a little unsure of this part of the proof. Doesn't the dominated convergence theorem only apply to sequences?
 A: The Dominated Convergence Theorem does apply to sequences. But you can apply it to any sequence $a_n$ with $a_n\to1$. So you are proving that $\varphi(a_n)\to\varphi(1)$ for every sequence $\{a_n\}$ with $\alpha_n\to1$, and $\varphi(a)=\int |f|^a$. This implies that $\varphi$ is continuous (continuity on a metric space can be tested on sequences). 
Now you stil need to deal with the case where $f$ is not bounded. 
A: From Tom Chen's comment, one can apply the sequential criterion for limits.  It suffices to show that for any sequence $(a_n)_n$ converging to $1$, $(||f||_{a_n})_n$ converges to $||f||_1$.
You don't need the assumption on the boundedness of $f$.  To apply the Dominated Convergence Theorem, you only need $|f|^a$ to be bounded by an $L^1([0,1])$ function.  Due to the condition at the beginning of the question, suppose that $f \in L^a([0,1])$ for some $a>1$.  Let $\delta \in (0,a-1)$, and $x \in (1-\delta,1+\delta)$.  You may try bounding $|f|^x$ by $1+|f|^a \in L^1([0,1])$.  (Why?)

 \begin{align} |f| &= |f| 1_{|f| \le 1} + |f| 1_{|f| > 1} \\ \forall x > 1, |f|^x &\le 1_{|f| \le 1} + |f|^x 1_{|f| > 1} \\ \forall x \in (1,a), |f|^x &\le 1_{|f| \le 1} + |f|^a 1_{|f| > 1} \le 1 + |f|^a \end{align}

Now, take $\delta \to 0$, so that $x \to 1$.  In your attempted proof, there should be an absolute sign for the pointwise convergence, so that $|f|^x$ is well-defined, and $|f|^x \to |f|$ pointwisely as $x \to 1$.  Therefore, as $n \to \infty$, $|f|^{a_n} \to |f|$ so $\int_{[0,1]} |f|^{a_n} \to \int_{[0,1]} |f|$ as $n \to \infty$, from which $||f||_{a_n} \to ||f||_1$.
