Show that $f$ is finite almost everywhere. Let $f:\mathbb{R} \rightarrow [0,\infty]$ be measurable so that $\int_{\mathbb{R}}^{}fdm<\infty$. Show that $f$ is finite almost everywhere.
Do I need to consider the subset $(0,1)\subset \mathbb{R}$ or by contradiction, let $E=\{x\in R,: |f|=\infty\}$, does it work?
Thanks in advance.
 A: Or, do the following. . First, note that
$$
\{x : f(x) = \infty\} = \bigcap_{n}\{x: f(x) >n\}
$$
hence, the set $\{x : f(x) =\infty\}$ is measurable. Next, suppose $\{x : f(x) = \infty\}$ has non-zero measure. Construct a function $g$ which is given as
$$
g(x) = \frac{\int f d\mu + 1}{\mu\left(\{x : f(x) = \infty\}\right)}\mathcal{X}_{f(x) = \infty}
$$
where $\mathcal{X}_E(x) = 1$ if $x \in E$ and $0$ otherwise. With this, note that $0 \leq g \leq f$ (indeed, if $f(x) \neq \infty$, then $g(x) = 0$, hence we trivially have $g(x) \leq f(x)$; otherwise, $f(x) = \infty$, again inequality holds trivially). Finally, from monotonicity of integral,
$$
\int g d\mu \leq \int f d\mu \implies \int fd\mu + 1 \leq \int f d \mu
$$
since $\int \mathcal{X}_{f(x) = \infty} d \mu = \mu\left(\left\{x: f(x) = \infty\right\}\right)$. Hence, we arrive at a contradiction.
A: Can be seen from the following argument (essentially Chebychev's inequality for the $L^1$-norm). For every $\lambda>0$, 
\begin{eqnarray*}
\infty>\|f\|_{L^1}
& = &
\int_{\{f>\lambda\}}f\; dm + \int_{\{f< \lambda\}}f\; dm
\\
&\geq & 
\int_{\{f>\lambda\}}f\; dm
\\
& \geq & 
\lambda\cdot m(\{f>\lambda\}). 
\end{eqnarray*} 
Therefore, for all $\lambda>0$, 
\begin{equation*}
m(\{f>\lambda\}) \leq \frac{\|f\|_{L^1}}{\lambda}. 
\end{equation*}
A: The problem can easily be reduced to the following: let $E \subseteq \mathbb{R}$ be a measurable set, and let $g: E \rightarrow [0,\infty]$ be the measurable function given by $f(x) = \infty$ for all $x \in E$.  If $m(E) > 0$, show that $\int\limits_E g dm = \infty$.
This follows directly from the definition of the integral.  A step function $\phi: E \rightarrow [0,\infty)$ is a function of the form
$$\sum\limits_{i=1}^t c_i \textrm{Char}(E_i)$$
where $E_1, ... , E_t$ are disjoint measurable sets whose union is $E$.  The integral $\int\limits_E \phi dm$ of such a step function is defined to be $\sum\limits_{i=1}^t c_i \, m(E_i)$, and $\int\limits_E g dm$ is defined to be the supremum of the integrals $\int\limits_E \phi dm$, as $\phi$ runs through all step functions for which $\phi \leq g$.  
Since $g(x) = \infty$ for all $x$, you have that $\phi \leq g$ for all step functions $\phi$.  So the problem becomes to show that the integrals $\int\limits_E \phi dm : \phi$ is a step function can become as big as you want.
