We assume that $\mu(X)>0$. Otherwise, the whole question does not make sense.
(a) (The solution has been pointed out by the eraldcoil. However,
it seems that the case $s=\infty$ is missing.) Let $r,s\in I_{f}$
with $r<s$. We go to show that $(r,s)\subseteq I_{f}$. Let $p\in(r,s)$
and let $E=\{x\mid f(x)>1\}$. We consider two cases.
Case 1: $s\neq\infty$. In this case, as pointed out by eraldcoil,
we have
\begin{eqnarray*}
\int f^{p} & = & \int_{E}f^{p}+\int_{E^{c}}f^{p}\\
& \leq & \int_{E}f^{s}+\int_{E^{c}}f^{r}\\
& < & \infty
\end{eqnarray*}
because $\int f^{s}<\infty$ and $\int f^{r}<\infty$.
Case 2: $s=\infty$. Note that $\mu(E)\leq\int_{E}f^{r}\leq\int f^{r}<\infty$.
$f\in L^{\infty}$ implies that there exists $M>0$ such that $\mu(\{x\mid f(x)>M\})=0$.
Therefore
\begin{eqnarray*}
\int f^{p} & = & \int_{E}f^{p}+\int_{E^{c}}f^{p}\\
& \leq & M^{p}\mu(E)+\int_{E^{c}}f^{r}\\
& < & \infty.
\end{eqnarray*}
This show that $p\in I_{f}$.
////////////////////////////////////////////////////////////////////////
(b) Let $r,s\in I_{f}$ with $r\leq s<\infty$. Let $\alpha\in(0,1)$
and define $p=\alpha r+(1-\alpha)s$. Define $p'=\frac{1}{\alpha}$,
$q'=\frac{1}{1-\alpha}$, then $p',q'\in(1,\infty)$ and $\frac{1}{p'}+\frac{1}{q'}=1$.
Now
\begin{eqnarray*}
||f||_{p}^{p} & = & \int f^{\alpha r}\cdot f^{(1-\alpha)s}\\
& \leq & ||f^{\alpha r}||_{p'}\cdot||f^{(1-\alpha)s}||_{q'}\\
& = & \left(\int f^{r}\right)^{\alpha}\cdot\left(\int f^{s}\right)^{(1-\alpha)}\\
& = & ||f||_{r}^{\alpha r}\cdot||f||_{s}^{(1-\alpha)s}.
\end{eqnarray*}
Note that (b) also gives a partial proof of (a) except for the case
that $s=\infty$.
Define $\phi:I_{f}\setminus\{\infty\}\rightarrow\mathbb{R}$ by $\phi(t)=\ln\left(||f||_{t}^{t}\right)$.
If $I_{f}\setminus\{\infty\}$ is a degenerated interval (i.e. empty
set or a singleton), there is nothing to prove. Suppose that $I_{f}\setminus\{\infty\}$
is a non-degenerated interval. Let $r,s\in I_{f}\setminus\{\infty\}$
with $r<s$. Let $\alpha\in(0,1)$. Define $p=\alpha r+(1-\alpha)s$.
We go to show that $\phi(p)\leq\alpha\phi(r)+(1-\alpha)\phi(s)$.
Firstly, note that since $f>0$ a.e., we have $0<\int f^{t}\leq\infty$ for
any $t\in[1,\infty)$. In particular, $||f||_{p}$, $||f||_{r}$,
$||f||_{s}$ are positive real numbers. Taking logarithm on both sides
of $||f||_{p}^{p}\leq||f||_{r}^{\alpha r}\cdot||f||_{s}^{(1-\alpha)s}$
yields $\phi(p)\leq\alpha\phi(r)+(1-\alpha)\phi(s).$ It follows that
$\phi$ is a convex function. Let $\psi:I_{f}^{o}\rightarrow\mathbb{R}$
be defined by $\psi(t)=||f||_{t}$. ($I_{f}^{o}$ denotes the interior
of the set $I_{f}$.) Since $\phi$ is convex, $\phi\bigg\vert_{I_{f}^{o}}$,
the restriction of $\phi$ on $I_{f}^{o}$ is continuous. Note that
for $t\in I_{f}^{o}$, $\psi(t)=\exp(\phi(t)/t)$ and that $t\geq1$,
so $\psi$ is a continuous function.
//////////////////////////////////////////////////////////////////////////
(x) Write $I_{f}=\langle a,b\rangle$ for some $a,b\in[1,\infty].$
Here $\langle$ may be $($ or $]$ and $\rangle$ may be $)$ or
$]$. Moreover, we assume that $a<b$ (otherwise, $I_f$ is a degenerated interval and there is nothing to prove).
Case 1: $I_{f}=\langle a,b]$. We go to show that $||f||_{p}\rightarrow||f||_{b}$
as $p\rightarrow b-$. Here, we need to distinguish the cases $b=\infty$,
$b<\infty$. Suppose that $b<\infty$. Let $(p_{n})_{n}$ be an arbitrary
sequence in $I_{f}$ such that $p_{n}\rightarrow b$ as $n\rightarrow\infty$.
Choose $c\in I_{f}$ such that $c\leq p_{n}$ for all $n$. Define
$E=\{x\mid f(x)>1\}$. Define $g=1_{E}f^{b}+1_{E^{c}}f^{c}$. Clearly
$f^{p_{n}}\leq g$ and $g$ is integrable. By Lebesgue dominated convergence
theorem, we have $\lim_{n}\int f^{p_{n}}=\int f^{b}$. Since
$(p_{n})_{n}$ is arbitary, by Heine theorem, it follows that $\lim_{p\rightarrow b-}\int f^{p}=\int f^{b}$.
Recall that the function $(0,\infty)\times(0,\infty)\rightarrow\mathbb{R}$,
$(x,y)\mapsto x^{y}$ is continuous (with respect to the product topology).
Now $\int f^{p}\rightarrow\int f^{b}$ and $\frac{1}{p}\rightarrow\frac{1}{b}$
with $(\int f^{b},\frac{1}{b})\in(0,\infty)\times(0,\infty)$. Therefore
$(\int f^{p})^{\frac{1}{p}}\rightarrow(\int f^{b})^{\frac{1}{b}}$.
That is, $||f||_{p}\rightarrow||f||_{b}$.
Then, we conside the case that $b=\infty$. Note that for any $p\in I_{f}$,
both the maps $\alpha\mapsto||\alpha f||_{p}$ and $\alpha\mapsto||\alpha f||_{\infty}$ are homogeneous in the sense that $||\alpha f||_{p}=\alpha||f||_{p}$
and $||\alpha f||_{\infty}=\alpha||f||_{\infty}$ for any $\alpha\geq0$.
Since $f>0$ a.e., we have that $M=||f||_{\infty}>0$. By replacing
$f$ with $2f/||f||_{\infty}$, without loss of generality, we may
assume that $M=2$. Fix $q\in I_{f}$ with $q<\infty$. Define $E=\{x\mid f(x)\geq1\}$.
We have that $\mu(E)\leq\int_{E}f^{q}<\infty$. Let $(p_{n})$ be
an arbitrary strictly increasing sequence in $I_{f}$ such that $p_{n}\rightarrow\infty$.
Without loss of generality, we may assume that $q<p_{n}$ for all
$n$. Define $A=\{x\mid f(x)>1.5\}$. Since $M=2$, we have that $\mu(A)>0$.
It follows that $\int_{E}f^{p_{n}}\geq\int_{A}f^{p_{n}}\geq(1.5)^{p_{n}}\mu(A)\rightarrow\infty$
as $n\rightarrow\infty$. On the other hand, the sequence of functions
$(1_{E^{c}}f^{p_{n}})$ is dominated by the integrable function $1_{E^{c}}f^{q}$
and $1_{E^{c}}f^{p_{n}}\rightarrow0$ pointwisely. By Lebesgue dominated
convergence theorem, we have $\int_{E^{c}}f^{p_{n}}\rightarrow0$
as $n\rightarrow\infty$. Choose $N$ such that $\int_{E^{c}}f^{p_{n}}\leq\int_{E}f^{p_{n}}$
whenever $n\geq N$. For any $n\geq N$, we have
\begin{eqnarray*}
\int f^{p_{n}} & = & \int_{E}f^{p_{n}}+\int_{E^{c}}f^{p_{n}}\\
& \leq & 2\int_{E}f^{p_{n}}\\
& \leq & 2M^{p_{n}}\mu(E).
\end{eqnarray*}
Hence $||f||_{p_{n}}\leq M\cdot(2\mu(E))^{\frac{1}{p_{n}}}$. It follows
that $\limsup_{n}||f||_{p_{n}}\leq\lim_{n}M\cdot(2\mu(E))^{\frac{1}{p_{n}}}=M$.
To show the reverse direction $\liminf_{n}||f||_{p_{n}}\geq M$, let
$B_{n}=\{x\mid f^{q}(x)\geq\frac{1}{n}\}$. Clearly $B_{1}\subseteq B_{2}\subseteq\ldots$
and $\cup_{n}B_{n}=\{x\mid f^{q}(x)>0\}$ whose complement has measure
zero. Since $f^{q}$ is integrable, we have $\mu(B_{n})<\infty$.
(In short, the support of $f$ is $\sigma$-finite) Let $\beta\in(0,1)$
be arbitrary. Define $C_{\beta}=\{x\mid f(x)>\beta M\}$. Since $\beta M$
is not an essential upper bound of $f$, $\mu(C_{\beta})>0$. By continuity
of measure, $0<\mu(C_{\beta})=\lim_{n}\mu(C_{\beta}\cap B_{n})$.
Hence, there exists $n_{0}$ such that $0<\mu(C_{\beta}\cap B_{n_{0}})<\infty$.
Denote $D=C_{\beta}\cap B_{n_{0}}$. We have
\begin{eqnarray*}
||f||_{p_{n}} & \geq & \left(\int_{D}f^{p_{n}}\right)^{\frac{1}{p_{n}}}\\
& \geq & \beta M\cdot\mu(D)^{\frac{1}{p_{n}}}.
\end{eqnarray*}
Letting $n\rightarrow\infty$, we have $\liminf||f||_{p_{n}}\geq\beta M$.
Finally, since $\beta$ is arbitrary, we have $\liminf||f||_{p_{n}}\geq M$.
This shows that $\lim_{n}||f||_{p_{n}}=M$. Since the sequence $(p_{n})$
is arbitrary, by Heine theorem, it follows that $\lim_{p\rightarrow\infty}||f||_{p}=||f||_{\infty}$.
Case 2: $I_{f}=[a,b\rangle$. This can be proved similarly to case
1, using Lebesgue dominated convergence theorem and the joint continuity
of the function $(x,y)\mapsto x^{y}$.
////////////////////////////////////////////////////////////////////////////////
(c) Let $r,s\in I_{f}$ with $r<s$. Let $p\in(r,s)$ be arbitrary.
Firstly, we consider the case that $s<\infty$. In this case, there
exists a unique $\alpha\in(0,1)$ such that $p=\alpha r+(1-\alpha)s$.
Since the two sides of the inequality are homogeneous, by replacing
$f$ with $f/||f||_{p}$, we may assume that $||f||_{p}=1$. By (b),
we have
\begin{eqnarray*}
1 & \leq & ||f||_{r}^{\alpha r}\cdot||f||_{s}^{(1-\alpha)s}\\
& = & \left(\int f^{r}\right)^{\alpha}\cdot\left(\int f^{s}\right)^{1-\alpha}.
\end{eqnarray*}
If $\int f^{r}=\int f^{s}$, then we have $\int f^{r}=\int f^{s}=1$
and hence $||f||_{r}=||f||_{s}=1$. If $\int f^{r}<\int f^{s}$, then
$\int f^{s}>1$. It follows that $\max(||f||_{r},||f||_{s})\geq||f||_{s}>1=||f||_{p}$.
If $\int f^{r}>\int f^{s}$, then $\int f^{r}>1$. We also have $\max(||f||_{r},||f||_{s})\geq||f||_{p}$.
Next, consider the case that $s=\infty$. Let $t>p$ be arbitrary.
Note that $r<p<t$ and $r,p,t\in I_{f}$. By the previous discussion,
we have $||f||_{p}\leq\max(||f||_{r},||f||_{t})$. By (x), $||f||_{t}\rightarrow||f||_{\infty}$
as $t\rightarrow\infty$. Since the function $x\mapsto\max(||f||_{r},x)$
is continuous, by letting $t\rightarrow\infty$, we have $||f||_{p}\leq\lim_{t\rightarrow\infty}\max(||f||_{r},||f||_{t})=\max(||f||_{r},||f||_{\infty})$.
The inclusion $L^{r}(X,\tau,\mu)\cap L^{p}(X,\tau,\mu)\subseteq\cap_{p\in(r,s)}L^{p}(X,\tau,\mu)$
follows trivially from the above inequality.
