Does $\int_{-\infty}^\infty f(x) dx < \infty$ where $f \ge 0$ implies $\sup_{x \in \mathbb R} f(x)<M $? Question. Does $\,\int_{-\infty}^\infty\, f(x)\, dx &lt; \infty,\,$ where $\,f \ge 0,\,$ imply that $\,\,\mathrm{ess}\sup_{x \in \mathbb R}\, f(x)&lt;\infty\,? $

My attempt

Since $f \ge 0$ and $\int_{-\infty}^\infty f(x) dx &lt; \infty$  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) &lt; 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?
No. A very simple counter-example can be constructed as follows.

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be defined by 
$$
f(x)=\begin{cases}
x^{-\frac{1}{2}}, &amp; \mbox{ if }x\in(0,1]\\
0, &amp; \mbox{ otherwise}
\end{cases}.
$$
Clearly $f$ is a non-negative Borel function and $\int f&lt;\infty$.
We go to show that $f$ is not essentially bounded. Let $M&gt;0$ be
arbitrary. Note that $\{x\mid|f(x)|&gt;M\}=(0,\frac{1}{M^{2}})$, which
has a positive Lebesgue measure. This shows that $|f|$ is not essentially
bounded.
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 $\int fdx&lt;+\infty$ but $f$ isn't bounded almost everywhere: let $k=\max_{x\in\mathbb R}\rho(x)$ and $M&gt;0$ generic then exists $i$ such that
$$
M&lt;k2^i
$$
Because $\rho$ is continue exists an open set $U\subseteq\mathbb R$ (its Lebesgue measure is strictly greater than $0$) such that for every $x\in U$
$$
M &lt; 2^i\rho(i+4^ix)\Rightarrow M&lt;f(x)
$$
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 $\epsilon&gt;0$ and increasing sequence $x_n$ such that $x_n\rightarrow +\infty$ and $f(x_n)&gt;\epsilon$.

From uniform continuity exists $\delta&gt;0$ such that for every $x, y\in\mathbb R$ such that $\lvert x-y\rvert &lt;\delta$ then $\lvert f(x)-f(y)\rvert&lt;\frac{\epsilon}{2}$. If $x\in ]x_n-\delta, x_n+\delta[$ then
$$
f(x)\geq f(x_n)-\lvert f(x)-f(x_n)\rvert&gt;\epsilon-\frac{\epsilon}{2}=\frac{\epsilon}{2}
$$
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
$$
\int^{+\infty}_{-\infty}f(x)dx\geq\int_Kf(x)dx=\begin{cases}
+\infty &amp; \text{ if }\lvert K\rvert &gt;0\\
0 &amp; \text{ if }\lvert K \rvert =0
\end{cases}
$$
so $f(x)&lt;+\infty$ for almost every $x\in\mathbb R$.