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

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.

• If $f$ takes value $\infty$ on a set of positive measure, then $f$ is not integrable. You may assume that $f$ is real-valued. Commented Oct 9, 2019 at 14:54
• That's what I said: that $f(x) = \infty$ almost nowhere, or in a set of measure $0$. Is the proof correct then? Commented Oct 9, 2019 at 14:58
• The dominated convergence theorem is applicable to "sequence" of functions only. Commented Oct 9, 2019 at 15:01
• I used the theorem for the sequence of functions $\{g_t\}$ Commented Oct 9, 2019 at 15:46
• $(g_t)$, indexed by positive real number $t$ is not a sequence. There are uncountably many measurable functions in $\{g_t \mid t>0\}$. Commented Oct 9, 2019 at 15:51

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 $$\lim_{n}t_{n}\mu(A_{n})=0$$.
\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 $$\mu(f=\infty)=0$$.