Condition for $L^r(\mu)=L^\infty(\mu)$ where $\mu(X)=1$ This is Exercise 3.5 in Rudin's Real and Complex Analysis. Suppose $\mu$ is a positive measure on a set $X$, with $\mu(X)=1$. Rudin askes: Under what conditions do the two spaces $L^r(\mu)$ and $L^s(\mu)$ contain the same functions, where $0<r<s\leq \infty$?
I am trying to show the case where $s=\infty$.
It is easy to show that $L^r(\mu)\supset L^\infty(\mu)$ using the Jensen's inequality. But I have no idea for finding a condtion on $X$ so that $L^r(\mu)=L^\infty(\mu)$. Any hints?
 A: I claim that if $\mu(X) = 1$ and $0 < r < s \leq \infty$, then the condition
$L^r(\mu) = L^s(\mu)$ is equivalent to the existence of $c > 0$ such that
$\mu(E) = 0$ or $\mu(E) \geq c$ for all measurable sets $E \subset X$.

"$\Rightarrow$":
Assume $L^r(\mu) = L^s(\mu)$.
Then the linear map
$$
  \Phi : L^r(\mu) \to L^s(\mu), f \mapsto f
$$
is well-defined.
Furthermore, if $f_n \to f$ in $L^r$ and $f_n \to g$ in $L^s$, then also $f_n \to g$ in $L^r$,
since (as you noted yourself) $L^s \hookrightarrow L^r$.
This implies that $\Phi$ has closed graph.
By the closed graph theorem (which also applies to quasi-Banach spaces; in fact, it applies to F-spaces), this means that
$\Phi$ is a bounded linear map, say $\| f \|_{L^s} \leq C \cdot \| f \|_{L^r}$ for all $f \in L^r$.
For $E \subset X$ measurable with $\mu(E) > 0$, this means
$$
  [\mu(E)]^{1/s}
  = \| 1_E \|_{L^s}
  \leq C \cdot \| 1_E \|_{L^r}
  =    C \cdot [\mu(E)]^{1/r} ,
$$
and hence $[\mu(E)]^{\frac{1}{r} - \frac{1}{s}} \geq C$,
so that finally $\mu(E) \geq C^{1/(\frac{1}{r} - \frac{1}{s})} =: c$,
where we used that $\frac{1}{r} - \frac{1}{s} > 0$.

"$\Leftarrow$":
Assume that $\mu(E) = 0$ or $\mu(E) \geq c$ for all measurable $E \subset X$.
It suffices to show (why?!) that $L^s (\mu) \hookrightarrow L^\infty(\mu)$.
In fact, I claim that if $f \in L^s$, then $\| f \|_{L^\infty} \leq c^{-1/s} \| f \|_{L^s}$.
Assume towards a contradiction that this is false; in particular, $\| f \|_{L^s} > 0$.
Since $\| f \|_{L^\infty} > c^{-1/s} \| f \|_{L^s}$, there is $\lambda > c^{-1/s} \| f \|_{L^s}$
such that $E := \{ x : |f(x)| \geq \lambda \}$ satisfies $\mu(E) > 0$ and hence $\mu(E) \geq c$.
This implies
$$
  \lambda \cdot c^{1/s}
  \leq \lambda \cdot [\mu(E)]^{1/s}
  = \| \lambda \cdot 1_{E} \|_{L^s}
  \leq \| f \|_{L^s}
$$
and hence $\lambda \leq c^{-1/s} \| f \|_{L^s}$, contradicting our choice of $\lambda$.
A: Let's say $(X,\mu)$ has the "small measure property" (SMP) if there exist pairwise disjoint $E_1,E_2,\dots\subset X$ such that $\mu(E_n)>0$ for all $n$ and $\mu(E_n)\to 0.$
Claim: Let $0<r<\infty.$ Then $L^r=L^\infty$ iff $X$ does not have the SMP.
Proof: $\implies$: Suppose, to reach a contradiction, $X$ has the SMP. Then we can choose pwdj $E_1,E_2,\dots$ such that $0<\mu(E_n)<2^{-n}.$ Define $f = \sum_{n=1}^{\infty} n\chi_{E_n}.$ Then
$$\int_X f^r\,d \mu = \sum_{n=1}^{\infty} n^r\mu(E_n)< \sum_{n=1}^{\infty} n^r2^{-n} < \infty.$$
So $f\in L^r.$ But $f\notin L^\infty$ because $f = n$ on $E_n, n=1,2,\dots,$ contradiction.
$\impliedby$: Let $f\in L^r.$ Set $E_n =\{n\le |f|<n+1\}.$ Note that the $E_n$ are pwdj. We have
$$\infty >\int_X|f|^r\,d\mu = \sum_{n=1}^{\infty}\int_{E_n}|f|^r\,d\mu \ge \sum_{n=1}^{\infty}n^r\mu(E_n).$$
Because the series on the right converges, we have $n^r\mu(E_n)\to 0,$ which certainly implies $m(E_n)\to 0.$ Since $X$ does not have the SMP, only finitely many $E_n$ have positive measure. Thus there is $N$ such that $\mu(E_n)=0$ for $n \ge N.$ This implies $|f|< N$ a.e., and thus $f\in L^\infty$ as desired.
