Prove a set is nowhere dense Fix $n \in \mathbb N$. Let $p \gt 1$.
Our whole space is $L^1([0,1])$.
So $L^p([0,1)]$ is contained in $L^1([0,1)]$.
Let $E_n=\{f\in L^1([0,1]) \mid \int_I |f| \le nm(I)^{1-\frac{1}{p}}$ for all intervals $ I \}$, where $m(I)$ is the lebesgue measure of $I$.
Prove that $E_n$ is nowhere dense in $L^1([0,1)]$.
This is an exercise in the Functional analysis written by Stein, exercise 9, on page 183.
Stein gives a hint that: consider $f_0+\epsilon g$, where $g(x)=x^{-(1-\delta)}$ with $0\lt \delta\lt 1-\frac{1}{p}$.
But $\int_{[0,1]} g$ diverges, right?
I consider this exercise for several hours, but I have no any idea.
Can someone help me? Thanks a lot.
 A: I think a slight modification in the function $f_0+\epsilon g$ is necessary for the argument to work. Suppose $E_n$ has an interior point $f_0$ and choose $\epsilon >0$ such that the closed ball of radius $\epsilon$ around $f_0$ is contained in $E_n$. Consider $h=f_0+\epsilon g s$ where $s(x)=1$ if $f_0 (x) \geq 0$ and $s(x)=-1$ if $f_0(x)<0, g(x)= \delta x^{-(1-\delta)}$.[$\delta$ as in the hint].  We have $f_0 +\epsilon g s $ has distance $\epsilon$ from $f_0$. Hence we must have $\int_a^{b}|f_0+\epsilon g s| \leq n(b-a)^{t}$ whenever $0<a<b<1$, where $t=1-\frac 1 p$. This gives $\int_a^{b}|f_0|+ \int_a^{b} \epsilon g \leq n(b-a)^{t}$ and hence $\int_a^{b} \epsilon g \leq n(b-a)^{t}$. But $\int_a^{b} g(x)dx=b^{\delta}-a^{\delta}$ so $\epsilon (b^{\delta}-a^{\delta}) \leq n(b-a)^{t}$. Put $a=0,b=1/k$ to get $k^{t-\delta} \leq \frac n \epsilon$ for all $k$. This is a contradiction because $\delta <t$. Remark: the function $s$ was introduced because $|f_0+\epsilon g|$ may not equal  $|f_0| + \epsilon |g|$
