Uniform convergence of $f_n$ implies convergence in $\mathcal{L}^1([0,1])$

Question

Let $f_n \in \mathcal{L}^1([0,1])$ for all $n\in \mathbb{N}$ then it follows that $f_n \rightarrow 0$ in $\mathcal{L}^1([0,1])$.

The proof goes as follows: Uniform convergence means i): $||f_n-0||_\infty \rightarrow 0$ and $f_n \rightarrow 0$ in $\mathcal{L}^1([0,1])$ means ii):$\int_0^1|f_n-0|d\lambda \rightarrow 0$.

So $\int_0^1|f_n-0|d\lambda \leq \int_0^1||f_n||_{\infty} d\lambda=||f_n||_\infty=||f_n-0||_\infty$ that confirms that ii) follows from i)

I am not sure if the equation $\int_0^1||f_n||_{\infty} d\lambda=||f_n||_\infty$ is right, but I don't see another way to proof it. How can I argue that?

Answer 1

For any constant function $k$, $\int_0^1k\,\mathrm d\lambda=k$. So, yes, your argument is correct.

Answer 2

Integral is linear, i.e. $$\int \alpha f d\lambda = \alpha \int f d\lambda$$ for any $\alpha\in\Bbb R$. In your case you also need to use $\int_0^1 1 d\lambda=1$.