$\lim_{n \to \infty}\int_{0}^{1}|f_{n}| = 0$ and $\limsup_{n \to \infty} f_{n}(x) = 1$ for all $x \in [0, 1]$?

Question

Is there a sequence $(f_{n})$ of $L^{1}$ functions such that $\lim_{n \to \infty}\int_{0}^{1}|f_{n}| = 0$ and $\limsup_{n \to \infty} f_{n}(x) = 1$ for all $x \in [0, 1]$?

I noticed that for every $\varepsilon > 0$ and every $x \in [0, 1]$ we have $$\int_{0}^{1}|f_{n}| < \varepsilon$$ for large $n$ and $$|\sup_{n \geq N}f_{n}(x) - 1| < \varepsilon$$ for large $N$, which is trivial, though.

Answer 1

Yes, the construction is essentially like this: Take the interval $[0,1]$ and call $f_1=1_{[0,1]}$. Then take the two halves of this interval and call $f_2=1_{[0,1/2]}$, $f_3=1_{[1/2,1]}$. Now take the halves of these intervals and ennumerate them as $f_4$, $f_5$, $f_6$, $f_7$. Then take the halves of these, etc. This sequence satisfies your conditions.

Answer 2

Choose $m$ and define $\phi_{i,m} = 1_{[{i \over m}, {i+1 \over m} ]}$ for $i = 0,...,m-1$.

Let $f_1 = \phi_{0,1}$, $f_2 = \phi_{0,2}, f_3 = \phi_{1,2}$,$\cdots$.