<s>[INST] Is continuous and integrable function bounded? I have a function $f: \mathbb R \rightarrow \mathbb R$ continuous and integrable on $\mathbb R$. Is $f$ bounded?
 [/INST] No! See this example here If $f\in L^1(\Bbb R,dx)$ then prove that for almost every $x\in\Bbb R$ $\lim\limits_{n\to \infty} f(nx) = 0.$
Where the function is a Polynomial on each interval $[n, n+\frac{1}{2^n}]$ and zero elsewhere.
Consider $f:\Bbb R\to \Bbb R$
\begin{align*}
f(x)= \begin{cases}
2^{n/2}P_n(x)& \text{if} ~~ n\le x\le n+\frac{1}{2^n} ~~\text{$n\in\mathbb{N}$}\\
0 &\text{ if}~~~j+\frac{1}{2^j}\le x\le j+1~~\text{ $j\in\mathbb{N}$}\\ f(-x)& \text{if }~~x  <0.
\end{cases}
\end{align*}
where $P_n(x)= -2^{2n+2}(x-n)^2+2^{n+2}(x-n)$. One can check that $f$ is an even and continuous function. since $P_n(n)= P_n(n+\frac{1}{2^n})=0$. Next observe that $f\in L^1(\Bbb R)$ since
\begin{align*}
\int_\mathbb{R} f(x)\,dx= 2\sum_{n=0}^{\infty}2^{n/2}\int_{n}^{n+\frac{1}{2^n}} P_n(x) dx 
\\= 2\sum_{n=0}^{\infty}-4\cdot 2^{n/2}
\left[ \frac{2^{2n}}{3}(x-n)^3-\frac{2^n}{2}(x-n)^2\right]_{n}^{n+\frac{1}{2^n}}\\
= \frac43\sum_{n=0}^{\infty} \frac{1}{2^{n/2}}<\infty. 
\end{align*}
Moreover, for every $n\in\mathbb{N}$ one has
\begin{align*}
f(n+\frac{1}{2^{n+1}}) = 2^{n/2} P_n ( n+\frac{1}{2^{n+1}}) = 2^{n/2}\to \infty 
\end{align*}
Therefore $f$ is unbounded.
Another example
\begin{align*}
f(x)= \begin{cases}
2^{\frac32(n+1)} x- n2^{\frac32(n+1)}& \text{if} ~~ n\le x\le n+\frac{1}{2^{n+1}} ~~\text{$n\in\mathbb{N}$}\\
-2^{\frac32(n+1)} x= n2^{\frac32(n+1)} +2^{\frac{n+3}{2}}& \text{if} ~~ n+\frac{1}{2^{n+1}}\le x\le n+\frac{1}{2^n} ~~\text{$n\in\mathbb{N}$}\\
0 &\text{ if}~~~j+\frac{1}{2^j}\le x\le j+1~~\text{$j\in\mathbb{N}$}\\ f(-x)& \text{if }~~x  <0.
\end{cases}
\end{align*}
It is easy to check that $f\in L^1(\Bbb R)\cap C(\Bbb R)$
\begin{align*}
&\int_\mathbb{R} f(x)\,dx=2\sum_{n=0}^{\infty} 2^{-\frac{n+1}{2}}<\infty\\
&f(n)= f(n+\frac{1}{2^{n}})=0\qquad f(n+\frac{1}{2^{n+1}}) = 2^{\frac{n+1}{2}}\to \infty 
\end{align*}
</s>