Proof of $\lim_{t\to\infty} t \mu(\{x: f(x) \geq t\}) = 0$ for $f \geq 0 $ integrable

Question

I am having some trouble proving this statement. This is what I have done, but I don't know if my reasoning is right or if I'm missing rigor.

For $f \geq 0 $ integrable on the measure space $(X, \mathcal{A},\mu)$, it is true that $$\lim_{t\to\infty} t \mu(\{x: f(x) \geq t\}) = 0$$

By the definition of the measure of a subset:

$$t \mu(\{x: f(x) \geq t\}) = t \int \chi_{\{x:f(x)\geq t\}} = \int t\chi_{\{x:f(x)\geq t\}} \leq \int f\chi_{\{x:f(x)\geq t\}} \leq \int f = \int|f| < \infty$$

Where the first inequality is due to the fact that on these sets, by definition $f(x) \geq t$ so $t \chi_{\{x:f(x)\geq t\}} \leq f\chi_{\{x:f(x)\geq t\}}$

Let $g_t(x)= t \chi_{\{x:f(x)\geq t\}} \to g(x) = \left\{\begin{array}{lr} \infty, & f(x)\geq\infty\to f(x)=\infty\\ 0, & f(x) < \infty \end{array}\right\}$

Now, since $f$ is integrable that means that $\mu(\{x: f(x)\notin \mathbb{R}\}) = 0$ otherwise we would find that $\int f = \int |f|=\infty$

Now $f$ is integrable, and it bounds $g_t$ for all $t$, and each $g_t$ is trivially measurable.

Therefore, we can use the Dominated Convergence Theorem and

$$\lim_{t\to \infty} t \mu(\{x:f(x)\geq t\}) = \lim_{t\to \infty}\int g_t = \int \lim_{t\to \infty} g_t = \int g=0$$ since $f(x) \notin \mathbb{R} $ almost nowhere.

The bit I am unsure of is the fact that an integrable non-negative $f$ implies that it is real-valued almost everywhere.

Answer 1

Let $(X,\mathcal{M},\mu)$ be a measure space. Let $f:X\rightarrow[0,\infty)$ be an integrable function. We go to prove that $\lim_{t\rightarrow\infty}t\mu\left(\left\{ x\in X\mid f(x)\geq t\right\} \right)=0$.

Proof: Let $(t_{n})$ be an arbitrary sequence with $0\leq t_{1}\leq t_{2}\leq\ldots$ and $t_{n}\rightarrow\infty$. By Heine theorem, it suffices to show that $\lim_{n\rightarrow\infty}t_{n}\mu\left(\left\{ x\in X\mid f(x)\geq t_{n}\right\} \right)=0$. (Remark: The axiom of choice is needed.) Denote $A_{n}=\left\{ x\in X\mid f(x)\geq t_{n}\right\} $. Note that \begin{eqnarray*} t_{n}\mu(A_{n}) & = & \int_{A_{n}}t_{n}d\mu\\ & \leq & \int_{A_{n}}fd\mu\\ & = & \int1_{A_{n}}fd\mu. \end{eqnarray*} Note that $|1_{A_{n}}f|\leq|f|$ and $f$ is integrable. By the Lebesgue Dominated Convergence Theorem, we have that $\lim_{n}\int1_{A_{n}}fd\mu=\int\lim_{n}1_{A_{n}}fd\mu=0$ because $1_{A_{n}}f\rightarrow0$ pointwisely.

It follows that $\lim_{n}t_{n}\mu(A_{n})=0$.

Answer 2

\begin{align*} \mu(f=\infty)&=\mu\left(\bigcap_{n}(f\geq n)\right)\\ &\leq\dfrac{1}{n}\int f, \end{align*} taking $n\rightarrow\infty$, we have $\mu(f=\infty)=0$.