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The post by Raziel shows that the answer to the original question is no. The OP then asked, in a comment to that post, if one one still conclude that $f(t)\asymp\frac1t$ (as $\to\infty$); as usual, $a\asymp b$ means here that $\limsup|\frac ab+\frac ba|<\infty$.

Let us show that the answer is still no. E.g., for natural $j$ let $t_j:=e^{j^2}$, \begin{equation} c_j:=\frac{\ln t_{j+1}-\ln t_j}{t_{j+1}-t_j}\sim\frac{2j}{t_{j+1}} \tag{1} \end{equation} (as $j\to\infty$), and \begin{equation} f(x):=c_j\quad\text{for}\quad x\in[t_j,t_{j+1}), \end{equation} with $f:=1$ on $[0,t_0)$. Let also $F(t):=\int_0^t f(x)\,dx$.

Then $f$ is nonincreasing, $0<f\le1$, $F(t_j)=1+\ln t_j\sim F(t_{j+1})$, whence $F(t)\sim\ln t$ (as $t\to\infty$), whereas $f(t_{j+1}-)=c_j$ is much greater than $\frac1{t_{j+1}}$, by (1). We also see that $f(t_j)=c_j$ is much less than $\frac1{t_j}$, again by (1).

The only condition missed here is the continuity of $f$. This omission is quite easy to fix by approximation.

The post by Raziel shows that the answer to the original question is no. The OP then asked, in a comment to that post, if one one still conclude that $f(t)\asymp\frac1t$ (as $\to\infty$); as usual, $a\asymp b$ means here that $\limsup|\frac ab+\frac ba|<\infty$.

Let us show that the answer is still no. E.g., for $j=0,1,\dots$ let $t_j:=e^{j^2}$, \begin{equation} c_j:=\frac{\ln t_{j+1}-\ln t_j}{t_{j+1}-t_j}\sim\frac{2j}{t_{j+1}} \tag{1} \end{equation} (as $j\to\infty$), and \begin{equation} f(x):=c_j\quad\text{for}\quad x\in[t_j,t_{j+1}), \end{equation} with $f:=c_0=\frac1{e-1}$ on $[0,t_0)$. Let also $F(t):=\int_0^t f(x)\,dx$.

Then $f$ is nonincreasing, $0<f\le1$, $F(t_j)=c_0+\ln t_j\sim c_0+\ln t_{j+1}=F(t_{j+1})$, whence $F(t)\sim\ln t$ (as $t\to\infty$), whereas $f(t_{j+1}-)=c_j$ is much greater than $\frac1{t_{j+1}}$, by (1). We also see that $f(t_j)=c_j$ is much less than $\frac1{t_j}$, again by (1).

The only condition missed here is the continuity of $f$, as $f$ is not left-continuous at $t_{j+1}$ for $j=0,1,\dots$. This omission is quite easy, but tedious, to fix by approximation. For instance, one can replace the above $f$ on every interval $[t_{j+1}-c_02^{-j},t_{j+1}]$ by the linear interpolation of $f$ on the same interval. Then instead of the value $c_0+\ln t_{j+1}$ of $F(t_{j+1})$ we will have $b_j+\ln t_{j+1}\sim c_0+\ln t_j=F(t_j)$ for some $b_j\in[0,c_0]$, and instead of $f(t_{j+1}-)=c_j$ being much greater than $\frac1{t_{j+1}}$, we will have that $f(t_{j+1}-c_0)=c_j$ is much greater than $\frac1{t_{j+1}-c_0}$.

The post by Raziel shows that the answer to the original question is no. The OP then asked, in a comment to that post, if one one still conclude that $f(t)\asymp\frac1t$ (as $\to\infty$); as usual, $a\asymp b$ means here that $\limsup|\frac ab+\frac ba|<\infty$.

Let us show that the answer is still no. E.g., for natural $j$ let $t_j:=e^{j^2}$, \begin{equation} c_j:=\frac{\ln t_{j+1}-\ln t_j}{t_{j+1}-t_j}\sim\frac{2j}{t_{j+1}} \tag{*} \end{equation} (as $j\to\infty$), and \begin{equation} f(x):=c_j\quad\text{for}\quad x\in[t_j,t_{j+1}), \end{equation} with $f:=1$ on $[0,t_0)$. Let also $F(t):=\int_0^t f(x)\,dx$.

Then $f$ is nonincreasing, $0<f\le1$, $F(t_j)=1+\ln t_j\sim F(t_{j+1})$, whence $F(t)\sim\ln t$ (as $t\to\infty$), whereas $f(t_{j+1}-)=c_j$ is much greater than $\frac1{t_{j+1}}$, by (*).

The only condition missed here is the continuity of $f$. This omission is quite easy to fix by approximation.

The post by Raziel shows that the answer to the original question is no. The OP then asked, in a comment to that post, if one one still conclude that $f(t)\asymp\frac1t$ (as $\to\infty$); as usual, $a\asymp b$ means here that $\limsup|\frac ab+\frac ba|<\infty$.

Let us show that the answer is still no. E.g., for natural $j$ let $t_j:=e^{j^2}$, \begin{equation} c_j:=\frac{\ln t_{j+1}-\ln t_j}{t_{j+1}-t_j}\sim\frac{2j}{t_{j+1}} \tag{1} \end{equation} (as $j\to\infty$), and \begin{equation} f(x):=c_j\quad\text{for}\quad x\in[t_j,t_{j+1}), \end{equation} with $f:=1$ on $[0,t_0)$. Let also $F(t):=\int_0^t f(x)\,dx$.

Then $f$ is nonincreasing, $0<f\le1$, $F(t_j)=1+\ln t_j\sim F(t_{j+1})$, whence $F(t)\sim\ln t$ (as $t\to\infty$), whereas $f(t_{j+1}-)=c_j$ is much greater than $\frac1{t_{j+1}}$, by (1). We also see that $f(t_j)=c_j$ is much less than $\frac1{t_j}$, again by (1).

The only condition missed here is the continuity of $f$. This omission is quite easy to fix by approximation.

The post by Raziel shows that the answer to the original question is no. The OP then asked, in a comment to that post, if one one still conclude that $f(t)\asymp\frac1t$ (as $\to\infty$); as usual, $a\asymp b$ means here that $\limsup|\frac ab+\frac ba|<\infty$.

Let us show that the answer is still no. E.g., for natural $j$ let $t_j:=e^{j^2}$, \begin{equation} c_j:=\frac{\ln t_{j+1}-\ln t_j}{t_{j+1}-t_j}\sim\frac{2j}{t_{j+1}} \tag{*} \end{equation} (as $j\to\infty$), and \begin{equation} f(x):=c_j\quad\text{for}\quad x\in[t_j,t_{j+1}), \end{equation} with $f:=1$ on $[0,t_0)$. Then $f$ is nondecreasing, $0<f\le1$, $F(t_j)=\ln t_j\sim F(t_{j+1})$, whence $F(t)\sim\ln t$ (as $t\to\infty$), whereas $f(t_{j+1}-)=c_j$ is much greater than $\frac1{t_{j+1}}$, by (*).

The only condition missed here is the continuity of $f$. This omission is quite easy to fix by approximation.

The post by Raziel shows that the answer to the original question is no. The OP then asked, in a comment to that post, if one one still conclude that $f(t)\asymp\frac1t$ (as $\to\infty$); as usual, $a\asymp b$ means here that $\limsup|\frac ab+\frac ba|<\infty$.

Let us show that the answer is still no. E.g., for natural $j$ let $t_j:=e^{j^2}$, \begin{equation} c_j:=\frac{\ln t_{j+1}-\ln t_j}{t_{j+1}-t_j}\sim\frac{2j}{t_{j+1}} \tag{*} \end{equation} (as $j\to\infty$), and \begin{equation} f(x):=c_j\quad\text{for}\quad x\in[t_j,t_{j+1}), \end{equation} with $f:=1$ on $[0,t_0)$. Let also $F(t):=\int_0^t f(x)\,dx$.

Then $f$ is nonincreasing, $0<f\le1$, $F(t_j)=1+\ln t_j\sim F(t_{j+1})$, whence $F(t)\sim\ln t$ (as $t\to\infty$), whereas $f(t_{j+1}-)=c_j$ is much greater than $\frac1{t_{j+1}}$, by (*).

The only condition missed here is the continuity of $f$. This omission is quite easy to fix by approximation.