Yes, the result is true. Below I give a proof, but I suspect a more direct argument is possible.
Let $(X,\mu)$ be a "nice" measure space, and $f:X\to\mathbb R$ be a measurable function such that $\int_X |f(x)|^p\ d\mu(x) <\infty$ for all $1\le p\le\infty$ (note that it is not necessary to include $\infty$, the argument will be the same either way).
For all $t\in[1,\infty)$, define the function
$$\phi_f : t\mapsto \|f\|_{L^t}^t := \int_X |f(x)|^t\ d\mu(x) $$
We will show that $\phi_f$ is continuous on its domain, which will imply the continuity of $t\mapsto \|f\|_{L^t}=\phi_f(t)^{1/t}$.
So let $t\in[1,\infty)$ and $(h_n)_{n\ge1}$ be a sequence of positive real numbers converging to $0$. For all $n$, we have
\begin{align*}
|\phi_f(t+h_n)-\phi_f(t)|&=\left|\int_X |f(x)|^{t+h_n}d\mu(x)-\int_X |f(x)|^{t}d\mu(x)\right|\\
&=\int_X \left[|f(x)|^{t+h_n} -|f(x)|^{t}\right] d\mu(x)\\
&=\int_X |f(x)|^{t}\left[|f(x)|^{h_n} -1\right]d\mu(x)\\
&=\int_{\{|f|\ge 1\}} |f(x)|^{t}\left[|f(x)|^{h_n} -1\right]d\mu(x) + \int_{\{|f|< 1\}} |f(x)|^{t}\left[|f(x)|^{h_n} -1\right]d\mu(x) \end{align*}
Now, by considering a subsequence if necessary, we have that $(h_n)$ is decreasing to $0$. Thus on the set $\{x\in X : |f(x)|\ge 1\} $ (respectively $\{x\in X :|f(x)|< 1\}$), we have that $\left(|f(x)|^{h_n}\right)_{n\ge 1} $ decreases (respectively increases) to $1$. Either way, it now follows from the monotone convergence theorem that
$$\lim_{n\to\infty}\int_{\{|f|\ge 1\}} |f(x)|^{t}\left[|f(x)|^{h_n} -1\right]d\mu(x) = 0 = \lim_{n\to\infty}\int_{\{|f|< 1\}} |f(x)|^{t}\left[|f(x)|^{h_n} -1\right]d\mu(x)$$
From which it finally follows that $\lim_{n\to\infty}|\phi_f(t+h_n)-\phi_f(t)| = 0 $. This shows that $\phi_f$ is right-continuous on $[1,\infty)$ (and thus continuous at $1$). A similar argument shows the left continuity of $\phi_f$ and thus we conclude that $\phi_f$ is continuous on $[1,\infty)$.
Some remarks :
- Note that I renamed $t\leftrightarrow p$ throughout the proof. This is purely cosmetic, and was not necessary at all. It's just easier for me to reason with $t$ as a variable rather than $p$, which usually is a fixed parameter (the same way that it would take me a minute to compute $\frac{d}{de}e^x$).
- It is also possible to show that $\lim_{p\to\infty} \|f\|_{L^p}=\|f\|_{L^{\infty}}$. See this post : Limit of $L^p$ norm
- As LL 3.14 said in a comment, some properties for $L^p$ norms with $p>1$ "break down" when $p=1$ or $p=\infty$, hence the need to separate cases sometimes. For a concrete example, it is known that if $1<p<\infty$, $L^p(X,\mu)$ is a reflexive space. But if $p=1$ or $p=\infty$, this is not true anymore, as shown here.
