A $\sigma$-finite Borel measure $\mu$ on $\mathbb R$ with $\mu(\mathbb R)=\infty$ s.t. $\int f \,d\mu=\int g \,d\mu \implies f=g$ pointwise? This is the final part of a problem on an old Analysis preliminary exam at my institution. We are given that $f,g\in L^1(\mathbb R,\mu)$ and $f\leq g$. 
For earlier parts of the problem, I've already showed that when $\mu$ is the Lebesgue measure, we have that $$\int_{\mathbb R} f\ d\mu=\int_{\mathbb R} g\ d\mu\implies f=g \quad\mathrm{a.e.}$$ I've also already showed that if we add the assumption that $f,g$ are continuous, then we can strengthen the above to $f=g$ (everywhere).
The final part of the problem wants a Borel measure as described in the title. I have a hard time thinking of possibilities to consider. The Lebesgue measure and the counting measure are the only two $\sigma-$finite Borel measures on $\mathbb R$ with $\mu(\mathbb R)=\infty$ that I can think of, and neither of them fits the bill. Any help would be appreciated.
Edit: Here is the exact text as it appeared on the preliminary exam.
Let $\lambda$ be the Lebesgue measure. Let $f,g\in L^1(\mathbb R,\lambda)$ satisfy $f\leq g$.
Part (a): If $\int_\mathbb{R}f d\lambda=\int_\mathbb{R}g d\lambda$, show that $f=g$ almost everywhere.
[Part (b) is not relevant to my question.]
Part (c): Describe a $\sigma-$finite Borel measure $\mu$ on $\mathbb R$ with $\mu(\mathbb R)=\infty$ with respect to which the conclusion of part (a) is true for all points (instead of almost everywhere).
 A: The question is saying (in less precise terms) this:
let $f, g \colon \mathbb R \longrightarrow \mathbb R$ be functions (so they're defined at all points $x \in \mathbb R$) such that $f \leq g$ (that is, $f(x) \leq g(x)$ for all $x \in \mathbb R$) and $f, g \in L^1(\lambda)$, where $\lambda$ is the Lebesgue measure on $\mathbb R$.
The question (c) is then: find a $\sigma$-finite Borel measure $\mu$ on $\mathbb R$ such that $\mu(\mathbb R) = \infty$ and 
$$
\int f \, d\mu = \int g \, d\mu \implies f(x) = g(x) \quad \forall x \in \mathbb R \tag{1}.
$$
We prove that no such $\mu$ exists by contradiction (filling in the details of the outline @ArunKumar gave). 
So, suppose that $\mu$ is a $\sigma$-finite Borel measure on $\mathbb R$ with $\mu(\mathbb R)= \infty$ which satisfies $(1)$. Fix $(c_0, x_0) \in \mathbb R^2$, and define $f, g, h \colon \mathbb R \longrightarrow \mathbb R$ by
\begin{align*}
f(x) &= c_0, \\
g(x) &= \begin{cases}
c_0, & x \neq x_0 \\
c_0 + 1, & x = x_0
\end{cases} \\
h(x) &= (g-f)(x) = \begin{cases}
0, & x \neq x_0 \\
1, & x = x_0
\end{cases}.
\end{align*}
These are all simple functions, so by the definition of the Lebesgue integral wrt $\mu$,
$$
\int(g-f) \, d\mu = \int h \, d\mu = 0 \cdot \mu\left( \mathbb R \backslash \{x_0\} \right) + 1 \cdot \mu(\{x_0\}) = \mu(\{x_0\})
$$
where we've used the fact that $\mu$ is a Borel measure. Since $f \neq g$ pointwise, the contrapositive of $(1)$ tells us that $\int f \, d\mu \neq \int g \, d\mu$, or
$$
0 \neq \int (g - f) \, d\mu = \int h \, d\mu = \mu(\{x_0\}),
$$
so $\{x_0\}$ has positive measure. Since $x_0 \in \mathbb R$ was arbitrary, we conclude that every nonempty subset of $\mathbb R$ has positive measure:
$$
\mathbb R \supseteq A \neq \varnothing \iff \mu(A) > 0. \tag{2}
$$
Now suppose that $\mathbb R$ is $\sigma$-finite:
$$
\mathbb R = \bigcup_{n \in \mathbb N} A_n, \qquad \mu(A_n) < \infty.
$$
For each $n \in \mathbb N$ we can write
$$
A_n = \bigcup_{k=1}^\infty \underbrace{\left\{ x \in \mathbb R : \tfrac{1}{k} \leq \mu(\{x\} \cap A_n) \right\}}_{A_{n,k}}.
$$
To confirm this, note that
\begin{align*}
x \in A_n &\implies \{x\} \cap A_n \neq \varnothing \\
&\implies \mu(\{x\} \cap A_n ) > 0 &&\text{by $(2)$} \\
&\implies \mu(\{x\} \cap A_n) \geq \tfrac{1}{k} \quad \text{for some } k \in \mathbb N && \text{Archimedean property of $\mathbb R$} \\
&\implies x \in A_{n,k}
\end{align*}
and
\begin{align*}
x \in A_{n,k} \text{ for some } k \in\mathbb N &\implies \mu(\{x\} \cap A_n) > 0 \\
&\implies \{x\} \cap A_n \neq \varnothing \\
&\implies x \in A_n.
\end{align*}
We claim that $A_{n,k}$ is finite, which will give us our contradiction, since we've expressed the uncountable set $\mathbb R$ as a countable union of finite sets $\bigcup_{n=1}^\infty \bigcup_{k=1}^\infty A_{n,k}$, which is countable. To see that $A_{n,k}$ is finite, suppose that $x_1, \dots, x_m$ are distinct points in $A_{n,k}$; then
\begin{align*}
\frac{1}{k} \leq \mu(\{x_i\} \cap A_n) &\implies \sum_{i=1}^m \frac{1}{k} \leq \sum_{i=1}^m \mu(\{x_i\} \cap A_n) = \mu\left( \bigcup_{i=1}^m \{x_i\} \cap A_n \right) \leq \mu(A_n) \\
&\implies m \leq k \mu(A_n) < \infty.
\end{align*}
(In particular, if we had chosen $m > k \mu(A_n)$, we'd get a contradiction.) Hence $A_{n,k}$ is finite, completing the proof.
A: It is not possible to find such a measure. If you want $\int f=\int g\implies f=g$, then you cannot have measure zero subsets, as else you can have two functions which are equal everywhere except on a measure zero set. Then they will have the same integral. In particular singleton sets will have non-zero measure. But then such a measure cannot be $\sigma$-finite.
Edit: I removed the second solution as I did not notice that the functions are $L^1$. But my above argument still holds, such a measure cannot have measure zero sets, I can choose the functions as above with $f\leq g$.
