For $f$ a continuous function on $[0,1]$ how to show that for any $\varepsilon>0$ there exists a function $$ \phi(x)=\sum_{k=1}^N \phi_j\chi_{[a_k,b_k)}(x) $$ such that $\phi(x)\geq f(x)$ for $x\in [0,1]$ and $\int_0^1 |f(x)-\phi(x)|dx<\varepsilon $?
Does this follows from the integrability of $f$?
Since $f$ is continuous on the compact interval $[0,1]$, it is uniformly continuous. Hence, for any $\epsilon > 0$ there is a $\delta > 0$ such that $|f(x)-f(y)| < \epsilon$ if $|x-y| < \delta$.
Choose a partition $(x_0,x_1, \ldots, x_n)$ of $[0,1]$ such that $\max(x_i - x_{i-1}) < \delta$.
Let
$$\phi_i = \sup\{f(x): x_{i-1} \leq x \leq x_i\}$$
and
$$\phi(x)=\sum_{i=1}^n \phi_i\chi_{[x_{i-1},x_i)}(x).$$
If $x \in [0,1]$, then for some $j$, $x \in [x_{j-1},x_j]$ and $\phi(x) = \phi_j \geq f(x).$
For any $\eta > 0$, there exists $z \in [x_{i-1},x_i]$ such that $\phi_i - \eta < f(z) \leq \phi_i$ and for any $x \in [x_{i-1},x_i]$ we have $\phi_i - f(x) \leq |f(x)-f(z)| + \eta< \epsilon + \eta.$ Since this is true for any $\eta > 0$ we have $\phi_i - f(x) < \epsilon$.
Furthermore,
$$\int_0^1|f(x) - \phi(x)|\, dx = \sum_{i=1}^n\int_{x_{i-1}}^{x_i}|f(x)-\phi_i|dx< \sum_{i=1}^n\epsilon(x_i-x_{i-1})=\epsilon$$