Is it true that if for $\varphi(p):=\int_E|f|^pd\mu$ we have $l=\lim_{x\uparrow b}\varphi(x)<+\infty$, then $\varphi(b)=l$?

Question

Let $(E,\mathcal{E},\mu)$ be a measured space and let $f:(E,\mathcal{E})\to(\mathbb{R},\mathfrak{B}(\mathbb{R}))$ be measurable. Define $\varphi(p):=\int_E|f|^pd\mu$ where $p>0$ and $I=\{p>0\ |\ \varphi(p)<+\infty\}$, then we know that $I$ is an interval and that $\varphi$ is continuous over $I$. But is it even true that if for $b\in]0,+\infty[$, if we don't know wether $b\in I$ or not, we have that $l=\lim_{x\uparrow b}\varphi(x)$ exits and is finite, then $\varphi(b)=l$ (and analogously for a decreasing limit)?

Answer 1

Assume that $\ell := \lim_{x \uparrow b} \varphi(x)$ exists and is finite, and pick some $p \in I$, $p<b$. Clearly,

$$|f(t)|^b \leq |f(t)|^p$$

for any $t \in E$ such that $|f(t)| \leq 1$, and so $$\int_{|f| \leq 1} |f(t)|^b \, d\mu(t) < \infty. \tag{1}$$ On the other hand, we have

$$|f(t)|^x \uparrow |f(t)|^b \quad \text{as $x \uparrow b$}$$

for any $t$ such that $|f(t)| \geq 1$. Applying the monotone convergence theorem, we find

$$\int_{|f| \geq 1} |f(t)|^b \, d\mu(t) = \sup_{x \uparrow b} \int_{|f| \geq 1} |f(t)|^x \, d\mu(t) \leq \sup_{x \uparrow b} \varphi(x) = \ell < \infty. \tag{2}$$

Combining $(1)$ and $(2)$ we find $\varphi(b)<\infty$. Now a straight-forward application of the dominated convergence theorem shows $\varphi(b) = \ell$.