Does $\int_{-\infty}^\infty f(x) dx < \infty$ where $f \ge 0$ implies $\sup_{x \in \mathbb R} f(x)Question. Does $\,\int_{-\infty}^\infty\, f(x)\, dx < \infty,\,$ where $\,f \ge 0,\,$ imply that $\,\,\mathrm{ess}\sup_{x \in \mathbb R}\, f(x)<\infty\,? $
My attempt
Since $f \ge 0$ and $\int_{-\infty}^\infty f(x) dx < \infty$  we get $\lim_{x \to \infty} f(x)=0$ and $\lim_{x \to -\infty} f(x)=0$ (we can show this easily by contradiction, the proof of which I skip). Since the limits exist can we say that the sequence is bounded and use this to conclude that $\sup_{x \in \mathbb{R}} f(x) < M $ (Lebesgue almost surely). Do we need continuity for this?
I am trying to show that if $f,g$ are non-negative  and the integral exists then $f\cdot g(x)$ exists for almost all $x$.
If not can you give a counterexample and explain the problem in my reasoning?
 A: No. A very simple counter-example can be constructed as follows.
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be defined by 
$$
f(x)=\begin{cases}
x^{-\frac{1}{2}}, & \mbox{ if }x\in(0,1]\\
0, & \mbox{ otherwise}
\end{cases}.
$$
Clearly $f$ is a non-negative Borel function and $\int f<\infty$.
We go to show that $f$ is not essentially bounded. Let $M>0$ be
arbitrary. Note that $\{x\mid|f(x)|>M\}=(0,\frac{1}{M^{2}})$, which
has a positive Lebesgue measure. This shows that $|f|$ is not essentially
bounded. 
A: No.
Consider 
$$
f(x)=\left\{
\begin{array}{ccc}
\cos \pi x & if & |x|\le 1/2,\\
0 & if & |x|>1/2. 
\end{array}
\right.
$$
Set
$$
g(x)=\sum_{n=1}^\infty n\,f\big(n^3(x-n)\big)
$$
Then
$$
g(x)\ge0, \qquad \int_{-\infty}^\infty g(x) \,dx
=\sum_{n=1}^\infty\int_{-\pi//n^3}^{\pi/n^3} 
n\cos(n^3\pi x)\,dx
=\sum_{n=1}^\infty \frac{2}{\pi n^2}=\frac{\pi}{3}$$ 
and 
$$
\sup_{x\in\mathbb R} g(x)=\infty.
$$
Note. If  $f(x)\ge 0$ and $\int_{-\infty}^\infty f(x)\,dx<\infty$, then this does not imply that $\lim_{x\to\infty} f(x)=0$.
A: Neither if $f$ is continue, let $\rho:\mathbb R\rightarrow [0, +\infty[$ a continue non identically zero function such that $\rho(x)\neq 0$ only inside $[-1, 1]$ and
$$
\int^{+\infty}_{-\infty}\rho(x)dx = 1
$$
Then let
$$
f(x)=\sum^{+\infty}_{i=1}2^i\rho\left(i+4^ix\right)
$$
$f$ is continuous and
$$
\int^{+\infty}_{-\infty}2^i\rho\left(i+4^ix\right)dx=\int^{4^{-i}(1-i)}_{4^{-i}(-1-i)}2^i\rho\left(i+4^ix\right)dx=2^i4^{-i}\int^{1}_{-1}\rho(y)dy\leq 2^{-i}
$$
then $\int fdx<+\infty$ but $f$ isn't bounded almost everywhere: let $k=\max_{x\in\mathbb R}\rho(x)$ and $M>0$ generic then exists $i$ such that
$$
M<k2^i
$$
Because $\rho$ is continue exists an open set $U\subseteq\mathbb R$ (its Lebesgue measure is strictly greater than $0$) such that for every $x\in U$
$$
M < 2^i\rho(i+4^ix)\Rightarrow M<f(x)
$$
and $f$ isn't almost everywhere bounded.
You need uniform continuity on $\mathbb R$ because statement
$$
\lim_{x\rightarrow \infty}f(x)=0
$$
is true if $f$ is uniformly continue.
Let $f$ uniformly continue on $\mathbb R$ and suppose exists $\epsilon>0$ and increasing sequence $x_n$ such that $x_n\rightarrow +\infty$ and $f(x_n)>\epsilon$.
From uniform continuity exists $\delta>0$ such that for every $x, y\in\mathbb R$ such that $\lvert x-y\rvert <\delta$ then $\lvert f(x)-f(y)\rvert<\frac{\epsilon}{2}$. If $x\in ]x_n-\delta, x_n+\delta[$ then
$$
f(x)\geq f(x_n)-\lvert f(x)-f(x_n)\rvert>\epsilon-\frac{\epsilon}{2}=\frac{\epsilon}{2}
$$
and exists an extract $n_k$ such that
$$
\int^{+\infty}_{-\infty}f(x)dx\geq\sum^{+\infty}_{k=0}\frac{\epsilon}{2}\left(x_{n_k}+\delta-x_{n_k}+\delta\right)=\sum^{+\infty}_{k=0}\epsilon\delta=+\infty
$$
absurd.
Your last assertion is true: let $f:\mathbb R\rightarrow [0, +\infty]$ such that integral is finite let $K=\{x\in\mathbb R : f(x)=+\infty\}$ then
$$
\int^{+\infty}_{-\infty}f(x)dx\geq\int_Kf(x)dx=\begin{cases}
+\infty & \text{ if }\lvert K\rvert >0\\
0 & \text{ if }\lvert K \rvert =0
\end{cases}
$$
so $f(x)<+\infty$ for almost every $x\in\mathbb R$.
