Does L1 and nonnegative imply bounded almost everywhere?

Question

Let $f:\mathbb{R}\longrightarrow\mathbb{R}$ a nonnegative function, such that $f\in L^1(\mathbb{R})$. Does this imply that $f$ is bounded almost everywhere?

Answer 1

Hint: What is $\int_0^1 x^\alpha\, dx$ (for a parameter $\alpha$)?

Answer 2

$f\left( x \right) = \frac{1}{{\sqrt x }}{1_{\left\langle 0 \right.,\left. 1 \right]}}\left( x \right) + \frac{1}{{{x^2}}}{1_{\left\langle {1, + \infty } \right\rangle }}\left( x \right) + {\delta _0}\left( x \right) + \frac{1}{{\sqrt { - x} }}{1_{\left[ { - 1} \right.,\left. 0 \right\rangle }} + \frac{1}{{{x^2}}}{1_{\left\langle { - \infty , - 1} \right\rangle }}\left( x \right)$ is defined on $\mathbb{R}$, is $f \in {L^1}\left( \mathbb{R} \right)$. Furthermore, $\lambda \left( {\left[ { - \frac{1}{n},\frac{1}{n}} \right]} \right) = \frac{2}{n} > 0$.

EDIT: Only defining it on ${\left\langle 0 \right.,\left. 1 \right]}$ or ${\left[ { - 1} \right.,\left. 0 \right\rangle }$ should be enough, I just woke up, so I wasn't thinking straight.