Prove that $\lim_{n \to \infty} \int_{\mathbb{R}} \min (f(x),f(x+1/n)) \, d\lambda (x) = \int_{\mathbb{R}} f \, d\lambda$? 
Let $f: \mathbb{R} \rightarrow [0,1]$ measurable and can be Lebesgue integrated. Is it true that
$$\lim_{n \rightarrow \infty} \int_{\mathbb{R}} \min (f(x),f(x+1/n)) \, d\lambda (x) = \int_{\mathbb{R}} f \, d\lambda \quad ?$$

I did not find a counterexample nor a proof for this. Could you help me?
EDIT: with @KaviRamaMurthy 's comment:
\begin{align*}
0
&\leq\int_{\mathbb{R}} f \, d\lambda - \lim_{n \to \infty} \int_{\mathbb{R}} \min (f(x),f(x+1/n)) \, d\lambda (x) \\
&=\lim_{n \to \infty}\int_{\mathbb{R}} f -\min (f(x),f(x+1/n)) \, d\lambda (x) \\
&\leq \lim_{n \to \infty}\int_{\mathbb{R}} \lvert f -\min (f(x),f(x+1/n)) \rvert \, d\lambda (x) \\
&\leq \lim_{n \to \infty}\int_{\mathbb{R}} \lvert f - f(x+1/n) \rvert \, d\lambda (x)  \\
&\to 0
\end{align*}
Is my proof correct? Or did I make some mistakes?
 A: We will write
$$ T_n f(x) = f(x+\tfrac{1}{n}) $$
for simplicity. Then by noting that $| (a \wedge b) - a | \leq |b - a|$, we get
$$ \| f \wedge (T_n f) - f \|_{L^1} \leq \| T_n f - f \|_{L^1}. $$
Now the $L^1$-continuity of the translation tells that $T_n f \to f$ in $L^1$, and so, $f \wedge (T_n f) \to f$ in $L^1$ as well.
A: Here is another approach without using the $L^{1}$ translation  fact, maybe somehow more complicated, but it depends:
By a step function $\varphi$ we mean that $\varphi$ is a finite linear combination of the characteristic functions of intervals.
Now by Lebesgue Dominated Convergence Theorem one has $\int\min(\varphi(x),\varphi(x+1/n))dx\rightarrow\int\varphi$ for any step function $\varphi$.
Use the well-known fact that step functions are $L^{1}$ dense, we pick a sequence $(\varphi_{N})$ of step functions such that $\varphi_{N}(x)\rightarrow f(x)$ a.e. and $\int|\varphi_{N}(x)-\varphi_{N-1}(x)|dx<2^{-N}$ for all $N=1,2,...$
By the inequality that $|\min(a,b)-a|\leq|b-a|$, we can further assume that $f,\varphi_{N}\geq 0$. The result will follow by triangle inequality once we consider $f=f^{+}-f^{-}$.
Now we see that
\begin{align*}
&\lim_{n\rightarrow\infty}\int\min(f(x),f(x+1/n))dx\\
&=\lim_{n\rightarrow\infty}\int\lim_{N\rightarrow\infty}\min(f(x),\varphi_{N}(x+1/n))dx\\
&=\lim_{n\rightarrow\infty}\lim_{N\rightarrow\infty}\int\min(f(x),\varphi_{N}(x+1/n))dx\\
&=\lim_{n\rightarrow\infty}\sum_{N=1}^{\infty}\int\bigg(\min(f(x),\varphi_{N}(x+1/n))-\min(f(x),\varphi_{N-1}(x+1/n))\bigg)dx,
\end{align*}
where we have used Lebesgue Dominated Convergence Theorem in the second equality by the observation that $0\leq\min(f(x),\varphi_{N}(x+1/n))\leq f(x)$, and in the last equality we set $\varphi_{0}=0$.
Recall the inequality that $|\min(a,b)-\min(a,c)|\leq|b-c|$, one has
\begin{align*}
&\left|\int\bigg(\min(f(x),\varphi_{N}(x+1/n))-\min(f(x),\varphi_{N-1}(x+1/n))\bigg)dx\right|\\
&\leq\int|\varphi_{N}(x+1/n)-\varphi_{N-1}(x+1/n)|dx\\
&=\int|\varphi_{N}(x)-\varphi_{N-1}(x)|dx\\
&<\dfrac{1}{2^{N}}.
\end{align*}
Since $\sum_{N}2^{-N}<\infty$, once again by Lebesgue Dominated Convergence Theorem we have
\begin{align*}
&\lim_{n\rightarrow\infty}\sum_{N=1}^{\infty}\int\bigg(\min(f(x),\varphi_{N}(x+1/n))-\min(f(x),\varphi_{N-1}(x+1/n))\bigg)dx\\
&=\sum_{N=1}^{\infty}\lim_{n\rightarrow\infty}\int\bigg(\min(f(x),\varphi_{N}(x+1/n))-\min(f(x),\varphi_{N-1}(x+1/n))\bigg)dx\\
&=\sum_{N=1}^{\infty}\int\bigg(\min(f(x),\varphi_{N}(x))-\min(f(x),\varphi_{N-1}(x))\bigg)dx\\
&=\lim_{N\rightarrow\infty}\int\min(f(x),\varphi_{N}(x))dx\\
&=\int\lim_{N\rightarrow\infty}\min(f(x),\varphi_{N}(x))dx\\
&=\int f(x)dx.
\end{align*}
